336. Palindrome Pairs 寻找回文对

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Given a list of unique words. Find all pairs of distinct indices (i, j) in the given list, so that the concatenation of the two words, i.e.words[i] + words[j] is a palindrome.

Example 1:
Given words = ["bat", "tab", "cat"]
Return [[0, 1], [1, 0]]
The palindromes are ["battab", "tabbat"]

Example 2:
Given words = ["abcd", "dcba", "lls", "s", "sssll"]
Return [[0, 1], [1, 0], [3, 2], [2, 4]]
The palindromes are ["dcbaabcd", "abcddcba", "slls", "llssssll"]

1.我的解答哈哈当然是超时的啦 sgin.....

就是遍历每个字符串，然后两两结合看是不是回文串

class Solution {public:    bool isPali(string s1, string s2){        string s = s1+s2;        for(int k = 0; k < s.size()/2; k++){            if(s[k] != s[s.size()-1-k])            return false;        }        return true;    }    vector<vector<int>> palindromePairs(vector<string>& words) {        int len = words.size();        vector<vector<int>>res;        if(len <= 0)        return res;        for(int i = 0; i < len; i++)        for(int j = i+1; j < len; j++){            if(isPali(words[i], words[j])){                vector<int> vec;                vec.push_back(i);                vec.push_back(j);                res.push_back(vec);            }            if(isPali(words[j], words[i])){                vector<int> vec;                vec.push_back(j);                vec.push_back(i);                res.push_back(vec);            }        }        return res;    }};

2.hash table

对每个字符串，对其倒置，对每个子串进行原始words中的匹配查找，若子串能找到且剩下一部分是回文，则可以组成回文串，加入结果集中。这里对左右两边的子串都进行判断。

注意：

1. 在第一个判断中加入j<len,是为了防止二次判断。

如：abcd, dcba

当对abcd字符串时，先reserve，然后从左边第0个开始分割，则dcba能找到；当分割到最后一个时，得到的又是dcba,则此时不加入结果集；因为当在对第二个字符串dcba开始处理时，会重复到。

class Solution {public:    bool isvalid(string s){        int i = 0, j = s.size()-1;        while(i < j){            if(s[i] != s[j])            return false;            i++;            j--;        }        return true;    }    vector<vector<int>> palindromePairs(vector<string>& words) {        vector<vector<int>>res;        map<string, int>map_str;        for(int i = 0; i < words.size(); i++)        map_str[words[i]] = i;                for(int j = 0; j < words.size(); j++){            reverse(words[j].begin(), words[j].end());            int len = words[j].size();            for(int k = 0; k <= len; k++){                string left = words[j].substr(0,k);                string right = words[j].substr(k);                if(map_str.count(left) && isvalid(right)&&(map_str[left] != j)&&(k < len))                    res.push_back(vector<int>{map_str[left], j});                if(map_str.count(right) && isvalid(left)&&(map_str[right] != j))                    res.push_back(vector<int>{j, map_str[right]});            }        }        return res;    }};

3.用trie树

下次再补

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