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  • 头条面试题

    #include <iostream>


    using namespace std;

    struct ListNode {
        int val;
        ListNode* next;
        ListNode(int x) : val(x), next(NULL) {}
    };

    ListNode* helper(ListNode* A, ListNode* B) {
        ListNode* ret = new ListNode(-1);
        ListNode* cur = ret;
        while (A && B) {
            if (A->val < B->val) {
                cur->next = A;
                A = A->next;
            }
            else {
                cur->next = B;
                B = B->next;
            }
            cur = cur->next;
        }
        while (A) {
            cur->next = A;
            A = A->next;
            cur = cur->next;
        }
        while (B) {
            cur->next = B;
            B = B->next;
            cur = cur->next;
        }
        return ret->next;
    }

    int main() {
        int n;
        cin >> n;
        auto A = new ListNode(-1);
        auto cur = A;
        for (int i = 0; i < n; i++) {
            int x;
            cin >> x;
            cur->next = new ListNode(x);
            cur = cur->next;
        }
        cin >> n;
        auto B = new ListNode(-1);
        cur = B;
        for (int i = 0; i < n; i++) {
            int y;
            cin >> y;
            cur->next = new ListNode(y);
            cur = cur->next;
        }
        auto ret = helper(A->next, B->next);
        while (ret) {
            cout << ret->val << " ";
            ret = ret->next;
        }
        cout << endl;
        return 0;
    }

    反转链表

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    #include <iostream>

    using namespace std;
    struct ListNode {
        int val;
        ListNode* next;
        ListNode(int x) : val(x), next(NULL) {}
    };


    ListNode* helper(ListNode* root) {
        ListNode* ret = new ListNode(-1);
        while (root) {
            auto temp = root->next;
            root->next = ret->next;
            ret->next = root;
            root = temp;
        }
        return ret->next;
    }

    int main() {
        int n;
        cin >> n;
        ListNode* A = new ListNode(-1);
        ListNode* cur = A;
        for (int i = 0; i < n; i++) {
            int x;
            cin >> x;
            cur->next = new ListNode(x);
            cur = cur->next;
        }

        auto ret = helper(A->next);
        while (ret) {
            cout << ret->val << " ";
            ret = ret->next;
        }
        return 0;
    }

    Partion and Reverse List

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    １->2->3->4->5->6->7
    空间复杂度Ｏ(1)转换成：
    1->7->2->6->3->5->4
    		 
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    #include <iostream>


    using namespace std;
    struct ListNode {
        int val;
        ListNode* next;
        ListNode(int x) : val(x), next(NULL) {}
    };

    ListNode* helper(ListNode* head) {
        if (!head || !head->next) return head;
        ListNode* fast = head, *slow = head;
        while (fast && fast->next) {
            fast = fast->next->next;
            slow = slow->next;
        }
        ListNode* head2 = slow->next;
        slow->next = NULL;
        ListNode* ret = new ListNode(-1);
        ListNode* cur = ret;
        while (head2) {
            auto temp = head2->next;
            head2->next = cur->next;
            cur->next = head2;
            cur = cur->next;
            head2 = temp;
        }
        head2 = ret->next;
        ret->next = NULL;

        cur = ret;
        while (head) {
            cur->next = head;
            head = head->next;
            cur = cur->next;
            if (head2) {
                cur->next = head2;
                head2 = head2->next;
                cur = cur->next;
            }
        }
        return ret->next;
    }


    int main() {
        int n;
        cin >> n;
        ListNode* head = new ListNode(-1);
        auto cur = head;
        for (int i = 0; i < n; i++) {
            int x;
            cin >> x;
            cur->next = new ListNode(x);
            cur = cur->next;
        }

        auto ret = helper(head->next);
        while (ret) {
            cout << ret->val << " ";
            ret = ret->next;
        }
        cout << endl;
        return 0;
    }

    单链表排序

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    // 快排版本
    #include <iostream>

    using namespace std;
    struct ListNode {
        int val;
        ListNode* next;
        ListNode(int x): val(x), next(NULL) {}
    };

    ListNode* helper(ListNode* head) {
        if (!head || !head->next) return head;
        ListNode* A = new ListNode(-1);
        ListNode* B = new ListNode(-1);
        ListNode* cur_A = A, *cur_B = B;
        auto base = head;
        head = head->next;
        while (head) {
            if (head->val < base->val) {
                cur_A->next = head;
                cur_A = head;
            }
            else {
                cur_B->next = head;
                cur_B = head;
            }
            head = head->next;
        }
        cur_A->next = NULL;
        cur_B->next = NULL;
        A->next = helper(A->next);
        B->next = helper(B->next);
        cur_A = A;
        while (cur_A->next) {
            cur_A = cur_A->next;
        }
        cur_A->next = base;
        base->next = B->next;
        return A->next;
    }


    int main() {
        int n;
        cin >> n;
        
        ListNode* head = new ListNode(-1);
        ListNode* cur = head;
        for (int i = 0; i < n; i++) {
            int x;
            cin >> x;
            cur->next = new ListNode(x);
            cur = cur->next;
        }
        
        auto ret = helper(head->next);
        while (ret) {
            cout << ret->val << " ";
            ret = ret->next;
        }
        cout << endl;
        return 0;
    }

    从一个数组里取m个数，能否和为n

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    // 类似于zeros and ones
    #include <iostream>
    #include <vector>

    using namespace std;

    bool helper(vector<int>& nums, int m, int s) {
        int n = nums.size();
        if (n < m) return false;
        vector<vector<bool>> dp(m + 1, vector<bool>(s + 1, false));
        dp[0][0] = true;

        for (auto num : nums) {
            for (int i = m; i >= 1; i --) {
                for (int j = s; j >= num; j--) {
                    dp[i][j] = dp[i][j] || dp[i - 1][j - num];
                }
            }
        }
        return dp[m][s];
    }

    int main() {
        int n;
        cin >> n;
        vector<int> nums(n, 0);
        for (int i = 0; i < n; i++) {
            cin >> nums[i];
        }

        int m, s;
        cin >> m >> s;
        cout << helper(nums, m, s) << endl;
        return 0;
    }

    最大子串和

    求出数组中最大的子串和，并求出子串

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    #include <iostream>
    #include <vector>
    #include <climits>

    using namespace std;

    vector<int> helper(vector<int>& nums) {
        int n = nums.size();
        int ans = 0;
        vector<int> temp;
        vector<int> cur;
        int ret = INT_MIN;
        for (auto i : nums) {
            ans += i;
            if (ans < 0) {
                ans = 0;
                cur = vector<int>({});
            }
            else {
                cur.push_back(i);
                if (ret < ans) {
                    ret = ans;
                    temp = cur;
                }
            }
        }
        return temp;
    }

    int main() {
        int n;
        cin >> n;
        vector<int> nums(n, 0);
        for (int i = 0; i < n; i++) {
            cin >> nums[i];
        }
        for (auto i : helper(nums)) {
            cout << i << endl;
        }
        return 0;
    }

    字符串拼接最大值

    一堆数字如123，324，56怎么拼接得到的值最大。

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    #include <iostream>
    #include <vector>
    #include <algorithm>

    using namespace std;

    bool cmp(const int& a, const int& b) {
        string x = to_string(a);
        string y = to_string(b);
        return x + y > y + x;
    }

    string helper(vector<int>& nums) {
        if (nums.empty()) return "";
        sort(nums.begin(), nums.end(), cmp);
        string ret;
        for (auto i : nums) {
            ret += to_string(i);
        }
        return ret;
    }

    int main() {
        int n;
        cin >> n;
        vector<int> nums(n, 0);
        for (int i = 0; i < n; i++) {
            cin >> nums[i];
        }
        cout << helper(nums) << endl;
        return 0;
    }

    左边最大值

    找出数组中每个数字左边部分（包括自己）最大的数字，然后返回结果数组

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    #include <iostream>
    #include <vector>
    #include <climits>
    #include <math.h>
    using namespace std;

    vector<int> helper(vector<int>& nums) {
        int n = nums.size();
        vector<int> ret(n, -1);
        int limit = INT_MIN;
        for (int i = 0; i < n; i++) {
            limit = max(limit, nums[i]);
            ret[i] = limit;
        }
        return ret;
    }

    int main() {
        int n;
        cin >> n;
        vector<int> nums(n, 0);
        for (int i = 0; i < n; i++) {
            cin >> nums[i];
        }
        for (auto i : helper(nums)) {
            cout << i << " ";
        }
        cout << endl;
        return 0;
    }

    Search in Rotated Sorted Array

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    #include <iostream>
    #include <vector>

    using namespace std;

    int helper(vector<int>& nums, int target) {
        int n = nums.size();
        int left = 0, right = n - 1;
        while (left <= right) {
            int mid = left + (right - left) / 2;
            if (nums[mid] == target) return mid;
            if (nums[mid] == nums[right]) {
                right --;
                continue;
            }
            if (nums[mid] > nums[right]) {
                if (nums[mid] > target && nums[right] < target) {
                    right = mid - 1;
                }
                else left = mid + 1;
            }
            else {
                if (nums[mid] < target && nums[right] > target) {
                    left = mid + 1;
                }
                else right = mid - 1;
            }
        }
        return -1;
    }

    int main() {
        int n;
        cin >> n;
        
        vector<int> nums(n, 0);
        for (int i = 0; i < n; i++) {
            cin >> nums[i];
        }

        int target;
        cin >> target;
        cout << helper(nums, target)<< endl;
        return 0;
    }

    数组变化

    一个数组，里面的元素全部初始为0，有以下两种操作：

    • 指定一个元素+1
    • 所有的*2

    问到达一个数组目标值得最小操作步数。

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    #include <iostream>
    #include <vector>
    #include <math.h>
    #include <climits>

    using namespace std;

    int helper(vector<int>& nums) {
        int ret = 0;
        int limit = INT_MAX;
        for (auto i : nums) {
            if (!i) continue;
            else {
                limit = min(limit, i);
                ret ++;
            }
        }
        int ans = 1;
        while (ans * 2 <= limit) ret ++, ans *= 2;
        for (auto i : nums) {
            if (i) ret += i - ans;
        }
        return ret;
    }

    int main() {
        int n;
        cin >> n;
        vector<int> nums(n, 0);
        for (int i = 0; i < n; i++) {
            cin >> nums[i];
        }
        cout << helper(nums) << endl;
        return 0;
    }

    顺时针打印数组

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    #include <iostream>
    #include <vector>
    #include <climits>
    using namespace std;

    vector<int> helper(vector<vector<int>>& nums) {
        int m = nums.size(), n = nums[0].size();
        int dir = 0;
        int a[4] = {0, 1, 0, -1}, b[4] = {1, 0, -1, 0};
        
        int x = 0, y = 0;
        vector<int> ret;
        while (ret.size() < m * n) {
            if (nums[x][y] != INT_MAX) {
                ret.push_back(nums[x][y]);
                nums[x][y] = INT_MAX;
            }
            int X = x + a[dir], Y = y + b[dir];
            if (X < m && X >= 0 && Y >= 0 && Y < n && nums[X][Y] != INT_MAX) {  
                x = X, y = Y;
            }
            else dir = (dir + 1) % 4;
        }
        return ret;
    }

    int main() {
        int m, n;
        cin >> m >> n;
        vector<vector<int>> nums(m, vector<int>(n, 0));
        for (int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                cin >> nums[i][j];
            }
        }
        for (auto i : helper(nums)) {
            cout << i << " ";
        } 
        cout << endl;
        return 0;
    }

    K个升序数组归并

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    #include <iostream>
    #include <vector>
    #include <utility>
    using namespace std;

    void heapfy(vector<pair<int, int>>& ans, vector<vector<int>>& nums, int index, int max) {
        int left = index * 2 + 1;
        int right = left + 1;
        int smallest = index;
        if (left < max && nums[ans[left].first][ans[left].second] < nums[ans[smallest].first][ans[smallest].second]) {
            smallest = left;
        }
        if (right < max && nums[ans[right].first][ans[right].second] < nums[ans[smallest].first][ans[smallest].second]) {
            smallest = right;
        }
        if (smallest != index) {
            swap(ans[index], ans[smallest]);
            heapfy(ans, nums, smallest, max);
        }
    }

    vector<int> helper(vector<vector<int>>& nums) {
        int n = nums.size();
        vector<pair<int, int>> ans;
        for (int i = 0; i < n; i++) {
            if (!nums[i].empty()) ans.push_back({i, 0});
        }
        vector<int> ret;
        while (!ans.empty()) {
            ret.push_back(nums[ans[0].first][ans[0].second]);
            if (ans[0].second + 1 < nums[ans[0].first].size()) {
                ans[0].second ++;
                heapfy(ans, nums, 0, ans.size());
            }
            else {
                ans[0] = ans.back();
                ans.pop_back();
                heapfy(ans, nums, 0, ans.size());
            }
        }
        return ret;
    }


    int main() {
        int k;
        cin >> k;
        vector<vector<int>> nums;
        for (int i = 0; i < k; i++) {
            int n;
            cin >> n;
            vector<int> ans(n, 0);
            for (int j = 0; j < n; j++) {
                cin >> ans[j];
            }
            nums.push_back(ans);
        }
        for (auto i : helper(nums)) {
            cout << i << " ";
        }
        cout << endl;
        return 0;
    }

    二叉树的最近公共祖先

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    #include <iostream>
    #include <vector>

    using namespace std;
    struct TreeNode {
        string val;
        TreeNode* left;
        TreeNode* right;
        TreeNode(string s) : val(s), left(NULL), right(NULL) {}
    };
    TreeNode* build(vector<string>& nums, int& index) {
        int n = nums.size();
        if (index >= n) return NULL;
        auto val = nums[index++];
        if (val == "#") return NULL;
        auto ret = new TreeNode(val);
        ret->left = build(nums, index);
        ret->right = build(nums, index);
        return ret;
    }

    string helper1(TreeNode* root, string a, string b, bool& m) {
        if (!root) return "";
        if (root->val == a || root->val == b) m = true;
        bool ml = false, mr = false;
        string l = helper1(root->left, a, b, ml);
        if (ml) {
            if (m) return root->val;
            m = true;
            string r = helper1(root->right, a, b, mr);
            if (mr) return root->val;
            else return l; 
        }
        else {
            string r = helper1(root->right, a, b, mr);
            if (m && mr) return root->val;
            if (mr) {
                m = mr;
                return r;
            }
            else return "";
        }
    }

    string helper(vector<string>& nums, string a, string b) {
        int index = 0;
        auto root = build(nums, index);
        bool m = false;
        return helper1(root, a, b, m);
    }


    int main() {
        int n;
        cin >> n;
        vector<string> nums;
        for (auto i = 0; i < n; i++) {
            string s;
            cin >> s;
            nums.push_back(s);
        }
        string a, b;
        cin >> a >> b;
        cout << helper(nums, a, b) << endl;
        return 0;
    }

    两个升序数组，查合并之后的总的中位数

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    #include <iostream>
    #include <vector>
    #include <math.h>
    using namespace std;
    typedef vector<int>::iterator Iter;

    int helper1(Iter a, int m, Iter b, int n, int k) {
        if (!n) return a[k - 1];
        if (!m) return b[k - 1];
        if (k == 1) return min(a[0], b[0]);

        int l = min(k / 2, m);
        int r = min(k / 2, n);
        if (a[l - 1] < b[r - 1]) {
            return helper1(a + l, m - l, b, n, k - l);
        }
        else {
            return helper1(a, m, b + r, n - r, k - r);
        }
    }

    double helper(vector<int>& A, vector<int>& B) {
        int n1 = A.size();
        int n2 = B.size();
        return (
            helper1(A.begin(), n1, B.begin(), n2, (n1 + n2 + 1) / 2) +
            helper1(A.begin(), n1, B.begin(), n2, (n1 + n2 + 2) / 2)
            ) / 2.0;
    }

    int main() {
        int n;
        cin >> n;
        vector<int> A(n, 0);
        for (int i = 0; i < n; i++) {
            cin >> A[i];
        }
        vector<int> B(n, 0);
        cin >> n;
        for (int i = 0; i < n; i++) {
            cin >> B[i];
        }
        cout << helper(A, B) << endl;
        return 0;
    }

    LRU

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    #include <iostream>
    #include <vector>
    #include <unordered_map>
    #include <list>
    #include <vector>
    #include <utility>

    using namespace std;

    class LRU {
    private:
        unordered_map<int, pair<int, list<int>::iterator>> nums;
        list<int> d;
        int cap;
        void touch(int key) {
            if (nums.find(key) == nums.end()) return;
            d.erase(nums[key].second);
            nums.erase(key);
        }
    public:
        LRU(int capacity): cap(capacity) {};
        void add(int key, int val) {
            touch(key);
            d.push_front(key);
            nums[key] = {val, d.begin()};
        }
        int get(int key) {
            if (nums.find(key) == nums.end()) return -1;
            int ret = nums[key].first;
            touch(key);
            d.push_front(key);
            nums[key] = {ret, d.begin()};
            return ret;
        }
    };
    int main() {
        auto cache = new LRU(2);
        cache->add(1, 1);
        cache->add(2, 2);
        cout << cache->get(1) << endl;       // returns 1
        
        return 0;
    }

    带重复的字符串全排列

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    #include <iostream>
    #include <vector>
    #include <algorithm>
    using namespace std;

    void dfs(vector<vector<int>>& ret, vector<int> nums, int index) {
        int n = nums.size();
        if (n == index) ret.push_back(nums);
        for (int i = index; i < n; i++) {
            if (i != index && nums[i] == nums[index]) continue;
            swap(nums[index], nums[i]);
            dfs(ret, nums, index + 1);
        }
    }

    vector<vector<int>> helper(vector<int>& nums) {
        vector<vector<int>> ret;
        sort(nums.begin(), nums.end());
        dfs(ret, nums, 0);
        return ret;
    }

    int main() {
        int n;
        cin >> n;
        vector<int> nums(n, 0);
        for (int i = 0; i < n; i++) {
            cin >> nums[i];
        }
        for (auto i : helper(nums)) {
            for (auto j : i) {
                cout << j << " ";
            }
            cout << endl;
        }
        return 0;
    }

    自然数排列

    0123456791011121314.. 自然数这样顺次排下去，给一个index，找出对应的数字是什么

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    #include <iostream>
    #include <math.h>
    using namespace std;
    char helper(int n) {
        if (!n) return '0';
        if (n < 10) return '1' + (n - 1);
        
        n -= 9;
        int cnt = 1;
        while (n > (pow(10, cnt + 1) - pow(10, cnt)) * (cnt + 1)) {
            n -= (pow(10, cnt + 1) - pow(10, cnt)) * (cnt + 1);
            cnt ++;
        }
        int ans = pow(10, cnt) + (n - 1) / (cnt + 1);
        return to_string(ans)[(n - 1) % (cnt + 1)];
    }
    int main() {
        for (int i = 0; i < 200; i++) {
            cout << helper(i);
        }
        cout << endl;
        return 0;
    }

    House Robber III

    Leetcode 337

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    class Solution {
    public:
        int helper(TreeNode* root, int& l, int& r) {
            if (!root) return 0;
            int ll = 0, lr = 0, rl = 0, rr = 0;
            l = max(0, helper(root->left, ll, lr));
            r = max(0, helper(root->right, rl, rr));
            return max(l + r, root->val + ll + lr + rl + rr);
        }
        int rob(TreeNode* root) {
            int l, r;
            return helper(root, l, r);
        }
    };

    Minimum Window Substring

    Leetcode 76

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    class Solution {
    public:
        string minWindow(string s, string t) {
            if (t.empty()) return "";
            unordered_map<char, int> m;
            for (auto i : t) {
                m[i] ++;
            }
            int ans = m.size();
            int n = s.size();
            int l = 0, r = 0;
            int len = INT_MAX;
            string ret;
            while (r <= n && l <= r) {
                if (ans > 0 && r < n) {
                    auto key = s[r++];
                    if (m.find(key) != m.end()) {
                        m[key] --;
                        if (m[key] == 0) ans--;
                    }
                    
                }
                else {
                    auto key = s[l++];
                    if (m.find(key) != m.end()) {
                        m[key] ++;
                        if (m[key] == 1) ans ++;
                    }
                    
                }
                if (!ans && r - l < len) {
                    len = r - l;
                    ret = s.substr(l, len);
                }
            }
            return ret;
        }
    };

    变色龙

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    Description
        在一个美丽的小岛上住着一群变色龙：其中有X只变色龙是红色的，Y只变色龙是绿色的，Z只变色龙是蓝色的。
        每个时刻会有两只不同颜色的变色龙相遇，相遇后他们会同时变成第三种颜色。比如，如果一只红色的变色龙和一只蓝色的变色龙相遇了，他们就会同时变成绿色的变色龙，如果一只绿色的变色龙和一只蓝色的变色龙相遇了，他们就会同时变成红色的变色龙，等等。
        那么最后是否有可能所有的变色龙都是同一种颜色呢？
     
    Input
        输入的第一行包含一个整数T (1 <= T <= 100)，表示接下来一共有T组测试数据。
        每组数据占一行，包含三个整数X, Y, Z (1 <= X, Y, Z <= 109)，含义同上。
     
    Output
        对于每组测试数据，如果最后有可能所有的变色龙都是同一种颜色，用一行输出“Yes”（不包括引号），否则输出“No”（不包括引号）。
     
    Sample Input
    4
    1 1 1
    1 2 3
    7 1 2
    3 7 5
    Sample Output
    Yes
    No
    Yes
    No
    HINT
    		 
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    #include <iostream>
    #include <math.h>
    using namespace std;

    string helper(int x, int y, int z) {
        if (x == y || y == z) return "YES";
        if (abs(x - y) % 3 == 0 || abs(x - z) % 3 == 0 || abs(y - z) % 3 == 0) return "YES";
        else return "NO";
    }

    int main() {
        int n;
        cin >> n;
        for (int i = 0; i < n; i++) {
            int x, y, z;
            cin >> x >> y >> z;
            cout << helper(x, y, z) << endl;
        }
        return 0;
    }

    平方根

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    #include <iostream>
    #include <math.h>
    #include <climits>
    using namespace std;

    int helper1(int n) {
        int ret = n;
        while (ret * ret > n) {
            ret = (ret + n / ret) / 2;
        }
        return ret;
    }
    int helper2(int n) {
        int left = 1, right = n;
        while (left <= right) {
            int mid = left + (right - left) / 2;
            if (mid + 1 > n / (mid + 1) && mid <= n / mid) return mid;
            if (mid < n / mid) left = mid + 1;
            else right = mid - 1;
        }
        return -1;
    }

    int main() {
        int n;
        cin >> n;
        cout << helper1(n) << endl;
        cout << helper2(n) << endl;
        return 0;
    }

    短网址系统

    设计一个短网址系统？短网址生成策略？短网址和长网址的映射关系如何表示？存网址的数据库表太大了怎么办？Sharding后如何分别以长网址或短网址为主key搜索？你觉得这个系统追求的是时间效率还是空间节省？那冗余存储的牺牲值不值得？

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    // 代码待补充

    大文件判断重复判断

    两个大文件，4g内存，判断两个文件里想同的url

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    /*
    三种思路供参考：
    1. 运用bitmap，类似于布隆过滤器
    2. 运用hash值，类似于布隆过滤器，有可能出现判断不准确的情况
    3. 分桶成小文件，然后查询过程中扫每个文件
    */

    无向图的最小环

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    输入：
    n个点，t条边
    输出：
    最短环的路径
    		 
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    #include <iostream>
    #include <vector>
    #include <utility>
    #include <unordered_set>
    #include <algorithm>
    using namespace std;


    vector<int> helper(vector<pair<int, int>>& nums, int n) {
        vector<unordered_set<int>> adj(n);
        for (auto i : nums) {
            if (i.first == i.second) continue;
            adj[i.first].insert(i.second);
            adj[i.second].insert(i.first);
        }

        vector<vector<int>> cur;
        cur.push_back({0});

        while (true) {
            vector<vector<int>> next;
            for (auto i : cur) {
                for (auto j : adj[i.back()]) {
                    if (i.size() > 2 && j == 0) {
                        i.push_back(0);
                        return i;
                    }
                    if (find(i.begin(), i.end(), j) == i.end()) {
                        auto temp = i;
                        temp.push_back(j);
                        next.push_back(temp);
                    }
                    
                }
            }
            if (next.empty()) break;
            cur = next;
        }
        return vector<int>();
    }

    int main() {
        int n;
        cin >> n;
        int t;
        cin >> t;
        vector<pair<int, int>> nums;
        for (int i = 0; i < t; i++) {
            int x, y;
            cin >> x >> y;
            nums.push_back({x, y});
        }


        for (auto i : helper(nums, n)) {
            cout << i << " ";
        }
        cout << endl;
        return 0;
    }
    /*
    6 7
    0 1
    1 2
    1 3
    2 3
    2 4
    4 5
    5 0
    */
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  • 原文地址：https://www.cnblogs.com/vector11248/p/11420341.html
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