LeetCode Weekly Contest 24

1, 543. Diameter of Binary Tree

维护左右子树的高度，同时更新结果，返回以该节点结束的最大长度。递归调用。

 /**
  * Definition for a binary tree node.
  * struct TreeNode {
  *     int val;
  *     TreeNode *left;
  *     TreeNode *right;
  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
  * };
  */
 class Solution {
 public:
 int res;
 int work(TreeNode* root) {
     if(!root) return ;
     int a = work(root->left), b = work(root->right);
     res = max(res, a + b + );
     return max(a, b) + ;
 }
 int diameterOfBinaryTree(TreeNode* root) {
     res = ;
     work(root);
     if(res >= ) res -= ;
     return res;
 }
 };

2. 538. Convert BST to Greater Tree

利用bst的有序性，一次遍历，同时维护后缀和，更新节点。递归调用。

 /**
  * Definition for a binary tree node.
  * struct TreeNode {
  *     int val;
  *     TreeNode *left;
  *     TreeNode *right;
  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
  * };
  */
 class Solution {
 public:
 int s;
 void  work(TreeNode* root) {
     if(!root) return;
     if(!root->left && !root->right) {
         root->val += s;
         s = root->val;
         return;
     }
     if(root->right) work(root->right);
     root->val += s;
     s = root->val;
     work(root->left);
 }
 TreeNode* convertBST(TreeNode* root) {
     s = ;
     work(root);
     return root;
 }
 };

3. 542. 01 Matrix

简单的bfs，初始为0的节点加入队列，然后更新相邻节点，直到最后的结果为空。

 int dx[] = {, -, , };
 int dy[] = {, , , -};
 class Solution {
 public:

 int n, m;
 bool check(int x, int y) {
     if(x >=  && x < n && y >=  && y < m) return ;
     return ;
 }
 vector<vector<int>> updateMatrix(vector<vector<int>>& matrix) {
     n = matrix.size();
     m = matrix[].size();
     typedef pair<int, int> pii;
     vector<vector<int>> res(n, vector<int>(m, INT_MAX));
     queue<pii> q;
     for (int i = ; i < n; i++) {
         for (int j = ; j < m; j++) {
             if(matrix[i][j] == ) {
                 res[i][j] = ;
                 q.push({i, j});
             }
         }
     }
     int x, y, v;
     while(!q.empty()) {
         x = q.front().first, y = q.front().second;
         q.pop();
         v = res[x][y];
         for (int i = ; i < ; i++) {
             int cx = x + dx[i], cy = y + dy[i];
             if(check(cx, cy)) {
                 if(res[cx][cy] > v + ) {
                     res[cx][cy] = v + ;
                     q.push({cx, cy});
                 }
             }
         }
     }
     return res;
 }
 };

4. 544. Output Contest Matches

一个队的rank在整个过程是不变的，都是第一个元素决定的，然后每次生成新节点，维护下顺序，第一个和最后一个合并，直到只剩一个为止。

 class Solution {
 public:

 string findContestMatch(int n) {
     vector<string> a;
     for (int i = ; i < n / ; i++) {
         stringstream ss;
         ss << "(" << i +  << "," << (n - i) << ")";
         a.push_back(ss.str());
     }
     vector<string> b;
     int sz = a.size();
     while(a.size() > ) {
         b.clear();
         sz = a.size();
         for (int i = ; i < sz / ; i++) {
             string t = "(" + a[i] + "," + a[sz - i - ] + ")";
             b.push_back(t);
         }
         swap(a, b);
     }
     return a[];
 }
 };