LeetCode Weekly Contest 21

1. 530. Minimum Absolute Difference in BST

最小的差一定发生在有序数组的相邻两个数之间，所以对每一个数，找他的前驱和后继，更新结果即可！再仔细一想，bst的中序遍历就是有序数组，每次记录上一个数字，求他们的差，更新结果即可！

class Solution {
public:
    int res;
    int x;
    void work(TreeNode* root) {
        if(!root) return;
        if(root->left) work(root->left);
        if(x != -1) {
            res = min(res, abs(x - root->val));
        }
        x = root->val;
        if(root->right) work(root->right);
    }
    int getMinimumDifference(TreeNode* root) {
        res = INT_MAX;
        x = -1;
        work(root);
        return res;
    }
};

2.  523. Continuous Subarray Sum

这题，我太坑了，错了4次，导致罚时非常多！

一看这题，不很简单么！1. 求连续的和为某一个数，用前缀和处理一下就好了， 2. 求是k的倍数，倍数很多怎么办，转化为求余数即可。（然后你就注意到：如果k=0，怎么解决，这是个大坑啊！）。

这题一定注意，长度要大于等于2啊！时刻牢记！

class Solution {
public:
    bool checkSubarraySum(vector<int>& nums, int k) {
        int n = nums.size();
        if(n < 2) return 0;
        if(k == 0) {
            for (int i = 0; i < n - 1; i++) {
                if(nums[i] + nums[i + 1] == 0) return 1;
            }
            return 0;
        }
        int s = 0;
        set<int> se;
        int c = 0;
        for (int x : nums) {
            s += x;
            c++;
            s %= k;
            if(s == 0 && c > 1) return 1;
            if(se.count(s)) return 1;
            se.insert(s);
        }
        return 0;
    }
};

我这个代码好像还不对，没有判读长度为2的情况！会不会重判，上面的代码，显然是错误的！

采用map记录一下下标！

class Solution {
public:
    bool checkSubarraySum(vector<int>& nums, int k) {
        int n = nums.size();
        if(n < 2) return 0;
        if(k == 0) {
            for (int i = 0; i < n - 1; i++) {
                if(nums[i] + nums[i + 1] == 0) return 1;
            }
            return 0;
        }
        int s = 0;
        set<int> se;
        map<int, int> ma;
        int c = 0;
        for (int x : nums) {
            s += x;
            c++;
            s %= k;
            if(s == 0 && c > 1) return 1;
            if(se.count(s) && (c - ma[s]) >= 2) return 1;
            se.insert(s);
            ma[s] = c;
        }
        return 0;
    }
};

3. 524. Longest Word in Dictionary through Deleting

这题就是暴力，先对字典排序，然后逐个比较，返回。

class Solution {
public:
string findLongestWord(string s, vector<string>& d) {
    int n = s.size();
    if(n == 0) return s;
    int m = d.size();
    if(m == 0) return "";
    sort(d.begin(), d.end(), [&](string x, string y) {
            if(x.size() == y.size()) return x < y;
            return x.size() > y.size();
        });
    for (string x : d) {
        if(x.size() > n) continue;
        int i, j;
        i = j = 0;
        int len = x.size();
        while(i < n && j < len) {
            if(s[i] == x[j]) {
                i++; j++;
            } else i++;
        }
        if(j == len) {
            return x;
        }
    }
    return "";
}
};

4. 529. Minesweeper

我就想，这次的最后一题分值有点小啊！原来是medium的！

理解题意，考虑各种情况，然后就是做一个bfs，记住判重和边界条件。

class Solution {
public:
int n, m;
bool check(int x, int y) {
    if(x >= 0 && x < n && y >= 0 && y < m) return 1;
    return 0;
}
vector<vector<char>> updateBoard(vector<vector<char>>& board, vector<int>& click) {
    n = board.size(), m = board[0].size();
    int x = click[0], y = click[1];
    if(board[x][y] == 'M') {
        board[x][y] = 'X';
        return board;
    }
    int s = 0;
    for (int i = -1; i <= 1; i++) {
        for (int j = -1; j <= 1; j++) {
            int cx = x + i, cy = y + j;
            if(!check(cx, cy)) continue;
            s += board[cx][cy] == 'M';
        }
    }
    if(s != 0) {
        board[x][y] = '0' + s;
        return board;
    }
    vector<vector<bool>> in(n, vector<bool>(m, 0));
    queue<pair<int, int>> q;
    q.push({x, y});
    while(!q.empty()) {
        auto t = q.front(); q.pop();
        x = t.first, y = t.second;
        s = 0;
        for (int i = -1; i <= 1; i++) {
            for (int j = -1; j <= 1; j++) {
                int cx = x + i, cy = y + j;
                if(!check(cx, cy)) continue;
                s += board[cx][cy] == 'M';
            }
        }
        if(s == 0) {
            board[x][y] = 'B';
            for (int i = -1; i <= 1; i++) {
                for (int j = -1; j <= 1; j++) {
                    int cx = x + i, cy = y + j;
                    if(!check(cx, cy)) continue;
                    if(board[cx][cy] == 'E' && !in[cx][cy]) {
                        q.push({cx, cy});
                        in[cx][cy] = 1;
                    }
                }
            }
        } else {
            board[x][y] = '0' + s;
        }
    }

    return board;
}
};