回文对

给定一组 互不相同 的单词， 找出所有 不同 的索引对 (i, j)，使得列表中的两个单词， words[i] + words[j] ，可拼接成回文串。

示例 1：

输入：words = ["abcd","dcba","lls","s","sssll"]
输出：[[0,1],[1,0],[3,2],[2,4]] 
解释：可拼接成的回文串为 ["dcbaabcd","abcddcba","slls","llssssll"]

示例 2：

输入：words = ["bat","tab","cat"]
输出：[[0,1],[1,0]] 
解释：可拼接成的回文串为 ["battab","tabbat"]

示例 3：

输入：words = ["a",""]
输出：[[0,1],[1,0]]

提示：

• 1 <= words.length <= 5000
• 0 <= words[i].length <= 300
• words[i] 由小写英文字母组成

#include <bits/stdc++.h>
using namespace std;
class Solution
{
public:
    vector<vector<int>> palindromePairs(vector<string> &words)
    {
        vector<vector<int>> res;
        unordered_map<string, int> m;
        set<int> s;
        for (int i = 0; i < words.size(); ++i)
        {
            m[words[i]] = i;
            s.insert(words[i].size());
        }
        for (int i = 0; i < words.size(); ++i)
        {
            string t = words[i];
            int len = t.size();
            reverse(t.begin(), t.end());
            if (m.count(t) && m[t] != i)
            {
                res.push_back({i, m[t]});
            }
            auto a = s.find(len);
            for (auto it = s.begin(); it != a; ++it)
            {
                int d = *it;
                if (isValid(t, 0, len - d - 1) && m.count(t.substr(len - d)))
                {
                    res.push_back({i, m[t.substr(len - d)]});
                }
                if (isValid(t, d, len - 1) && m.count(t.substr(0, d)))
                {
                    res.push_back({m[t.substr(0, d)], i});
                }
            }
        }
        return res;
    }
    bool isValid(string t, int left, int right)
    {
        while (left < right)
        {
            if (t[left++] != t[right--])
                return false;
        }
        return true;
    }
};