1. 557. Reverse Words in a String III
分割字符串，翻转。
1 class Solution { 2 public: 3 string reverseWords(string s) { 4 int n = s.size(); 5 if(n < 2) return s; 6 string res; 7 int i = 0; 8 n += 1; 9 s += " "; 10 int last = 0; 11 while(i < n) { 12 last = i; 13 while(i < n && s[i] != ' ') i++; 14 string t = s.substr(last, i - last); 15 reverse(t.begin(), t.end()); 16 res += t + " "; 17 i++; 18 } 19 return res.substr(0, n - 1); 20 } 21 };
2. 554. Brick Wall
这刚好是以前做的hihocode的一道题目：https://hihocoder.com/problemset/problem/1494
就是统计边界的最大次数，然后n- max求出最小的穿过次数。
1 class Solution { 2 public: 3 int leastBricks(vector<vector<int>>& wall) { 4 int n = wall.size(); 5 if(n == 0) return 0; 6 map<int, int> ma; 7 int res = 0; 8 for (vector<int> &it:wall) { 9 int m = it.size(); 10 int s = 0; 11 for (int i = 0; i < m - 1; i++) { 12 s += it[i]; 13 ma[s]++; 14 res = max(res, ma[s]); 15 } 16 } 17 return n - res; 18 } 19 };
3. 556. Next Greater Element III
要求下一个比较大的，可以采用next_permutation来做，然后判断下是不是真的比n大，然后还要check是否超过int的表示范围。
1 class Solution { 2 public: 3 int nextGreaterElement(int n) { 4 if(n < 10) return -1; 5 vector<int> v; 6 int t = n; 7 while(t > 0) { 8 v.push_back(t % 10); 9 t /= 10; 10 } 11 int sz = v.size(); 12 //cout << sz << endl; 13 if(sz == 1) return -1; 14 reverse(v.begin(), v.end()); 15 if(next_permutation(v.begin(), v.end())) { 16 long long res = 0; 17 for (int t : v) { 18 res = res * 10 + t; 19 } 20 if(res > INT_MAX) 21 return -1; 22 return res; 23 } else 24 return -1; 25 } 26 };
4. 549. Binary Tree Longest Consecutive Sequence II
这个题跟求二叉树的最长路径的方法是一致的，维护信息，以及更新结果。
我当时写的有点复杂，只需要维护每个点的最长上升和下降长度就行。注意：题目要求必须连续。
1 /** 2 * Definition for a binary tree node. 3 * struct TreeNode { 4 * int val; 5 * TreeNode *left; 6 * TreeNode *right; 7 * TreeNode(int x) : val(x), left(NULL), right(NULL) {} 8 * }; 9 */ 10 class Solution { 11 public: 12 int res; 13 typedef pair<pair<int, int>, pair<int, int>> pp; 14 typedef pair<int, int> pii; 15 pair<pair<int, int>, pair<int, int>> work(TreeNode* root) { 16 if(!root) { 17 return {{0,0}, {0, 0} }; 18 } 19 if(!root->left && !root->right) { 20 int v = root->val; 21 res = max(res, 1); 22 return {{1, v}, {1, v} }; 23 } 24 int v = root->val; 25 pp m1 = work(root->left), m2 = work(root->right); 26 int a1 = m1.first.first, a2 = m1.first.second, b1 = m1.second.first, b2 = m1.second.second; 27 int ta1 = m2.first.first, ta2 = m2.first.second, tb1 = m2.second.first, tb2 = m2.second.second; 28 int ra1 = 1, ra2 = v, rb1 = 1, rb2 = v; 29 if(v == a2 + 1) { 30 ra1 = a1 + 1; 31 } 32 if(v == ta2 + 1) { 33 ra1 = max(ra1, ta1 + 1); 34 } 35 if(v == b2 - 1) { 36 rb1 = b1 + 1; 37 } 38 if(v == tb2 - 1) { 39 rb1 = max(rb1, tb1 + 1); 40 } 41 if(v == a2 + 1 && v == tb2 - 1) { 42 res = max(res, 1 + a1 + tb1); 43 } 44 if(v == ta2 + 1 && v == b2 - 1) { 45 res = max(res, 1 + ta1 + b1); 46 } 47 res = max(res, max(ra1, rb1)); 48 return {{ra1, ra2}, {rb1, rb2} }; 49 50 } 51 int longestConsecutive(TreeNode* root) { 52 if(!root) return 0; 53 res = 0; 54 work(root); 55 return res; 56 } 57 };