leetcode-----第 26 场双周赛

1446. 连续字符

• 链接：https://leetcode-cn.com/problems/consecutive-characters/

代码

class Solution {
public:
    int maxPower(string s) {
        int tmp = 1, ans = 1;
        for (int i = 1; i < s.size(); ++i) {
            if (s[i] == s[i - 1]) ans = max(ans, ++tmp);
            else tmp = 1;
        }
        return ans;
    }
};

1447. 最简分数

• 链接：https://leetcode-cn.com/problems/simplified-fractions/

代码

class Solution {
public:
    int gcd(int a, int b) {
        if (b == 0) return a;
        return gcd(b, a % b);
    }
    vector<string> simplifiedFractions(int n) {
        vector<string> ans;
        for (int i = 2; i <= n; ++i) {
            for (int j = 1; j < i; ++j) {
                if (gcd(i, j) == 1) {
                    ans.push_back(to_string(j) + "/" + to_string(i));
                }
            }
        }
        return ans;
    }
};

1448. 统计二叉树中好节点的数目

• 链接：https://leetcode-cn.com/problems/count-good-nodes-in-binary-tree/

代码

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    int ans = 0;
    void dfs(TreeNode* node, int cmax) {
        if (!node) return;

        if (node->val >= cmax) {
            cmax = node->val;
            ans++;
        }

        dfs(node->left, cmax);
        dfs(node->right, cmax);
    }

    int goodNodes(TreeNode* root) {
        dfs(root, -1e9);
        return ans;
    }
};

1449. 数位成本和为目标值的最大数字

• 链接：https://leetcode-cn.com/problems/form-largest-integer-with-digits-that-add-up-to-target/

思路

  完全背包问题求解
  f[i, j] = max{f(i - 1, j), f(i, j - cost[j]) + 1};

代码

class Solution {
public:
    string largestNumber(vector<int>& cost, int target) {
        string ans;
        vector<vector<int>> f(10, vector<int>(target + 1));

        for (int i = 1; i <= target; ++i) f[0][i] = -1e8;

        for (int i = 1; i <= 9; ++i) {
            for (int j = 0; j <= target; ++j) {
                f[i][j] = f[i - 1][j];
                if (j >= cost[i - 1]) f[i][j] = max(f[i][j], f[i][j - cost[i - 1]] + 1);
            }
        }

        if (f[9][target] < 1) return "0";

        for (int i = 9, j = target; i ; i--) {
            while (j >= cost[i - 1] && f[i][j] == f[i][j - cost[i - 1]] + 1) {
                ans += to_string(i);
                j -= cost[i - 1];
            }
        }
        return ans;
    }
};