336. Palindrome Pairs

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.

【思路】

x和y能形成回文的就2种情况： 1、xright | xleft | y              2、 y | xright | xleft

第一种情况 遍历x 直到发现 到某一个节点  xleft 是回文的 而 xright ==y   也就是前一部分能在hashtable中找到 而后一部分是回文的

第二种情况相反       前一部分是回文的 而后一部分能在hashtable 中找到

注意点 在hashtable中查找时 erase掉当前x

先把所有单词 放入hashtable  对每一个单词从左到右 遍历一遍，

class Solution {public:    vector<vector<int>> palindromePairs(vector<string>& words) {        vector<vector<int>> ans;        unordered_map<string,int>dic;        int size =words.size();        for(int i =0l;i<size;i++){            dic[words[i]]=i;        }        for(int i =0;i<size;i++){            int len =words[i].length();            dic.erase(words[i]);            for(int j =0;j<=len;j++){                string subl= words[i].substr(0,j);                string subr =words[i].substr(j);                string revl,revr;                revl =revl.assign(subl.rbegin(),subl.rend());                revr =revr.assign(subr.rbegin(),subr.rend());                if (j != len && isPalindrome(subr) && dic.count(revl))                       ans.push_back(vector<int> {i, dic[revl]});                  if (isPalindrome(subl) && dic.count(revr))                      ans.push_back(vector<int> {dic[revr], i});              }            dic[words[i]]=i;        }        return ans;    }     bool isPalindrome(string s) {          int left = 0, right = s.length()-1;          while (left < right) {              if (s[left] != s[right])                  return false;              left++;              right--;          }          return true;      }  };