336. Palindrome Pairs

问题描述：

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"]

Credits:
Special thanks to @dietpepsi for adding this problem and creating all test cases.

解题思路：

我一开始的想法是把所有单词根据其最后一个字符存入hash表，构成一个<char, vector<string>>的map

然后再遍历每一个单词，根据他们第一个单词在hash表中寻找，构成新的字符串，来判断是否为回文，若是则要加入。

但是这个方法耗时很大，若n为数组长度，k为字符串平聚长度。 应为n*(n-1)*2k

还有一种方法是O(n*k²) n 为数组长度，而k为字符串的平均长度。

这里需要额外注意的是！

字符串数组中！有没有肯恩含有空数组！！！

代码：

我的：

class Solution {
public:
    vector<vector<int>> palindromePairs(vector<string>& words) {
        vector<vector<int>> ret;
        unordered_map<string, int> m;
        unordered_map<char, vector<string>> endWith;
        bool emptyExist = false;
        
        for(int i = 0; i < words.size(); i++){
            m[words[i]] = i;
            if(words[i] != "")
                endWith[words[i].back()].push_back(words[i]);
            else
                emptyExist = true;
        }
        
        for(int i = 0; i < words.size(); i++){
            if(emptyExist && words[i] != ""){
                if(isPalindrome(words[i])){
                    ret.push_back({m[""], i});
                    ret.push_back({i, m[""]});
                }
            }
            for(auto str : endWith[words[i][0]]){
                string newS = words[i] + str;
                if(m[str] != i && isPalindrome(newS)){
                    ret.push_back({i, m[str]});
                }
            }
        }
        return ret;
    }
    bool isPalindrome(string s){
        int l = 0, r = s.size()-1;
        while(l < r){
            if(s[l++] != s[r--]) return false;
        }
        return true;
    }
};

快的：

class Solution {
public:
    vector<vector<int>> palindromePairs(vector<string>& words) {
        vector<vector<int>> ret;
        unordered_map<string, int> dict;
        
        for(int i = 0; i < words.size(); i++){
            string w = words[i];
            reverse(w.begin(), w.end());
            dict[w] = i;
        }
        
        if(dict.count("") != 0){
            for(int i = 0; i < words.size(); i++){
                if(words[i] != ""){
                    if(isPalindrome(words[i]))
                        ret.push_back({dict[""], i});
                }
            }
        }
        
        for(int i = 0; i < words.size(); i++){
            for(int j = 0; j < words[i].size(); j++){
                string left = words[i].substr(0, j);
                string right = words[i].substr(j, words[i].size() - j);
                if(dict.count(left) != 0 && isPalindrome(right) && dict[left] != i)
                    ret.push_back({i, dict[left]});
                if(dict.count(right) != 0 && isPalindrome(left) && dict[right] != i)
                    ret.push_back({dict[right], i});
            }
        }
        return ret;
    }
    bool isPalindrome(string s){
        int l = 0, r = s.size()-1;
        while(l < r){
            if(s[l++] != s[r--]) return false;
        }
        return true;
    }
};