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 0;
    int a = work(root->left), b = work(root->right);
    res = max(res, a + b + 1);
    return max(a, b) + 1;
}
int diameterOfBinaryTree(TreeNode* root) {
    res = 0;
    work(root);
    if(res >= 1) res -= 1;
    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 = 0;
    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[] = {1, -1, 0, 0};
int dy[] = {0, 0, 1, -1};
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<int>> updateMatrix(vector<vector<int>>& matrix) {
    n = matrix.size();
    m = matrix[0].size();
    typedef pair<int, int> pii;
    vector<vector<int>> res(n, vector<int>(m, INT_MAX));
    queue<pii> q;
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < m; j++) {
            if(matrix[i][j] == 0) {
                res[i][j] = 0;
                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 = 0; i < 4; i++) {
            int cx = x + dx[i], cy = y + dy[i];
            if(check(cx, cy)) {
                if(res[cx][cy] > v + 1) {
                    res[cx][cy] = v + 1;
                    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 = 0; i < n / 2; i++) {
        stringstream ss;
        ss << "(" << i + 1 << "," << (n - i) << ")";
        a.push_back(ss.str());
    }
    vector<string> b;
    int sz = a.size();
    while(a.size() > 1) {
        b.clear();
        sz = a.size();
        for (int i = 0; i < sz / 2; i++) {
            string t = "(" + a[i] + "," + a[sz - i - 1] + ")";
            b.push_back(t);
        }
        swap(a, b);
    }
    return a[0];
}
};