LeetCode 336回文对(回文串+哈希)

题目大意：给出一组字符串，输出所有的两个字符串拼接能够形成回文串的组合
样例输入：[“abcd”,“dcba”,“lls”,“s”,“sssll”]
样例输出：[[0,1],[1,0],[3,2],[2,4]]
思路：枚举所有可能的拼接情况，判断每次拼接是否是回文串，TLE。
暴力代码：

class Solution {
    
    
private:
    bool isHuiWen(string a){
    
    
        int i = 0, j = a.size()-1;
        while (i < j){
    
    
            if (a[i] != a[j]) return false;
            i++;j--;
        }
        return true;
    }
public:
    vector<vector<int>> palindromePairs(vector<string>& words) {
    
    
        vector<vector<int>> ans;
        int len = words.size();
        for (int i = 0; i < len; i++){
    
    
            for (int j = i+1; j < len; j++){
    
    
                if (isHuiWen(words[i]+words[j])){
    
    
                    vector<int> t;t.push_back(i);t.push_back(j);
                    ans.push_back(t);
                }
                if (isHuiWen(words[j]+words[i])){
    
    
                    vector<int> t;t.push_back(j);t.push_back(i);
                    ans.push_back(t);
                }
            }
        }
        return ans;
    }
};

优化思路：假设拼接的字符串为$a，b$，不妨设$len(a)>=len(b)$，如果$a$和$b$拼接后的字符串是回文串，那么字符串$a$的前$len(b)$个字符串是字符串$b$的转置，且$a$剩余的字符串也是回文串。
实现方法：使用$map<string,int>$记录所有字符串和下标，对于每一个字符串，判断其每一个子串是否是回文串，剩余部分的转置是否与其他字符串相同
AC代码：

class Solution {
    
    
private:
    map<string, int> mp;
    bool isHuiWen(string s, int a, int b){
    
    
        int i = a, j = b;
        while (i < j){
    
    
            if (s[i] != s[j]) return false;
            i++;j--;
        }
        return true;
    }
public:
    vector<vector<int>> palindromePairs(vector<string>& words) {
    
    
        vector<vector<int>> ans;
        int len = words.size();
        for (int i = 0; i < len; i++){
    
    
            mp[words[i]] = i;
        }
        for (int i = 0; i < len; i++){
    
    
            int m = words[i].size();
            for (int j = 0; j <= m; j++){
    
    
                if (isHuiWen(words[i], j, m-1)){
    
    
                    string t = words[i].substr(0, j);
                    reverse(t.begin(), t.end());
                    if (mp.count(t) && mp[t] != i){
    
    
                        ans.push_back(vector<int>{
    
    i, mp[t]});
                    }
                }
                if (j && isHuiWen(words[i], 0, j-1)){
    
    
                    string t = words[i].substr(j, m);
                    reverse(t.begin(), t.end());
                    if (mp.count(t) && mp[t] != i){
    
    
                        ans.push_back(vector<int>{
    
    mp[t], i});
                    }
                }
            }
        }
        return ans;
    }
};