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 != -) {
             res = min(res, abs(x - root->val));
         }
         x = root->val;
         if(root->right) work(root->right);
     }
     int getMinimumDifference(TreeNode* root) {
         res = INT_MAX;
         x = -;
         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 < ) return ;
         if(k == ) {
             for (int i = ; i < n - ; i++) {
                 if(nums[i] + nums[i + ] == ) return ;
             }
             return ;
         }
         int s = ;
         set<int> se;
         int c = ;
         for (int x : nums) {
             s += x;
             c++;
             s %= k;
             if(s ==  && c > ) return ;
             if(se.count(s)) return ;
             se.insert(s);
         }
         return ;
     }
 };

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 < ) return ;
         if(k == ) {
             for (int i = ; i < n - ; i++) {
                 if(nums[i] + nums[i + ] == ) return ;
             }
             return ;
         }
         int s = ;
         set<int> se;
         map<int, int> ma;
         int c = ;
         for (int x : nums) {
             s += x;
             c++;
             s %= k;
             if(s ==  && c > ) return ;
             if(se.count(s) && (c - ma[s]) >= ) return ;
             se.insert(s);
             ma[s] = c;
         }
         return ;
     }
 };

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 == ) return s;
     int m = d.size();
     if(m == ) 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 = ;
         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 >=  && x < n && y >=  && y < m) return ;
     return ;
 }
 vector<vector<char>> updateBoard(vector<vector<char>>& board, vector<int>& click) {
     n = board.size(), m = board[].size();
     int x = click[], y = click[];
     if(board[x][y] == 'M') {
         board[x][y] = 'X';
         return board;
     }
     int s = ;
     for (int i = -; i <= ; i++) {
         for (int j = -; j <= ; j++) {
             int cx = x + i, cy = y + j;
             if(!check(cx, cy)) continue;
             s += board[cx][cy] == 'M';
         }
     }
     if(s != ) {
         board[x][y] = '' + s;
         return board;
     }
     vector<vector<bool>> in(n, vector<bool>(m, ));
     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 = ;
         for (int i = -; i <= ; i++) {
             for (int j = -; j <= ; j++) {
                 int cx = x + i, cy = y + j;
                 if(!check(cx, cy)) continue;
                 s += board[cx][cy] == 'M';
             }
         }
         if(s == ) {
             board[x][y] = 'B';
             for (int i = -; i <= ; i++) {
                 for (int j = -; j <= ; 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] = ;
                     }
                 }
             }
         } else {
             board[x][y] = '' + s;
         }
     }

     return board;
 }
 };