LeetCode 热题 HOT 100 第十七天

Leetcode394. 字符串解码

思路：递归

字母直接加入到$res$中，碰到数字求出数字长度（重复的次数），递归后半段，将右边重复出现的字符接在$res$后面，返回$res$

class Solution {
public:
    string decodeString(string s) {
        int u = 0;
        return dfs(s, u);
    }

    string dfs(string& s, int& u) {
        string res;
        while (u < s.size() && s[u] != ']') {
            if (s[u] >= 'a' && s[u] <= 'z' || s[u] >= 'A' && s[u] <= 'Z') res += s[u ++ ];
            if (s[u] >= '0' && s[u] <= '9') {
                int k = u;
                while (s[u] >= '0' && s[u] <= '9') u ++ ;
                int x = stoi(s.substr(k, u - k));
                u = u + 1;
                string y = dfs(s, u);
                while (x -- ) res += y;
                u ++ ;    
            }
        }
        return res;
    }
};

Leetcode399. 除法求值

思路：哈希+floyd

class Solution {
public:
    vector<double> calcEquation(vector<vector<string>>& equations, vector<double>& values, vector<vector<string>>& queries) {
        unordered_set<string> S;
        unordered_map<string, unordered_map<string, double>> d;
        for (int i = 0; i < equations.size(); i ++ ) {
            auto a = equations[i][0], b = equations[i][1];
            auto c = values[i];
            d[a][b] = c, d[b][a] = 1 / c;
            S.insert(a), S.insert(b);
        }

        for (auto k : S)
            for (auto i : S)
                for (auto j : S)
                    if (d[i][k] && d[k][j])
                        d[i][j] = d[i][k] * d[k][j];
        vector<double> res;
        for (auto & t : queries) {
            auto a = t[0], b = t[1];
            if (d[a][b]) res.push_back(d[a][b]);
            else res.push_back(-1.0);
        }
        return res;
    }
};

Leetcode406. 根据身高重建队列

在这里插入代码片

Leetcode416. 分割等和子集

class Solution {
public:
    bool canPartition(vector<int>& nums) {
        int n = nums.size(), s = 0;
        for (auto& t : nums) s += t;
        if (s & 1) return false;
        s /= 2;
        vector<int> f(s + 1);
        f[0] = 1;
        for (auto& t : nums)
            for (int j = s; j >= t; j -- )
                f[j] |= f[j - t];
        return f[s];
    }
};

Leetcode437. 路径总和 III

思路：前缀和

/**
 * 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:
    unordered_map<int, int> cnt;
    int res = 0;
    int pathSum(TreeNode* root, int sum) {
        cnt[0] ++ ;
        dfs(root, sum, 0);
        return res;
    }
    void dfs(TreeNode* root, int sum, int cur){
        if (!root) return ;
        cur += root->val;
        res += cnt[cur - sum];
        cnt[cur] ++ ;
        dfs(root->left, sum, cur), dfs(root->right, sum, cur);
        cnt[cur] -- ;
    }
};