第 199 场周赛

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Words count in article: 480 | Reading time ≈ 2

重新排列字符串

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class Solution {
public:
    string restoreString(string s, vector<int>& indices) {
        string res = s;
        for(int i = 0; i < indices.size(); ++i)
            res[indices[i]] = s[i];
        return res;
    }
};

灯泡开关 IV

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class Solution {
public:
    int minFlips(string target) {
        int flag = true;
        int res = 0;
        for(int i=0; i<target.size(); i++) {
            if(flag && target[i] == '1') {
                flag = false;
                res++;
            } else if(flag == false && target[i] == '0') {
                flag = true;
                res++;
            }
        }
        return res;
    }
};

好叶子节点对的数量

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/**
 * 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;
    vector<int> dfs(TreeNode* root, int distance) {
        if (root->left == nullptr&&root->right == nullptr)
            return{ 1 };
        vector<int>v1, v2,v3;
        if (root->left)
            v1 = dfs(root->left, distance);
        if (root->right)
            v2 = dfs(root->right, distance);
        for (int i : v1)
            for (int j : v2)
                if (i + j <= distance)
                    ans++;
        for (int i : v1)
            v3.push_back(i + 1);
        for (int j : v2)
            v3.push_back(j + 1);
        return v3;
    }
    int countPairs(TreeNode* root, int distance) {
        ans = 0;
        dfs(root, distance);
        return ans;
    }
};

压缩字符串 II

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class Solution {
public:
    int dp[105][105];
    int n;
    int dfs(int st, int k, string& s) {
        if (st == n) return 0;
        int& r = dp[st][k];
        if (r != -1) return r;
        if (k) r = dfs(st + 1, k - 1, s); // case1: 删除当前位置字符

        // case2: 保留当前字符，并枚举该字符连续的长度（所以遇到不同字符必须删除，才能维护其连续性）
        int c = 1; // 记录当前字符的连续个数
        for (int i = st + 1; i <= n; i++) {
            int t = dfs(i, k, s) + 1 + (c >= 100 ? 3 : c >= 10 ? 2 : c > 1 ? 1 : 0);
            if (r == -1 || r > t) r = t;
            if (i == n) break;
            if (s[i] == s[st]) c++; // 相同, 个数加1
            else if (k-- <= 0) break; // 否则删除不同的字符
        }
        return r;
    }
    int getLengthOfOptimalCompression(string s, int k) {
        n = s.size();
        memset(dp, -1, sizeof dp);
        dfs(0, k, s);
        return dp[0][k];
    }
};
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