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能形成迴文的就2種情況： 1、xright | xleft | y              2、 y | xright | xleft

第一種情況 遍歷x 直到發現 到某一個節點  xleft 是迴文的 而 xright ==y   也就是前一部分能在hashtable中找到 而後一部分是迴文的

第二種情況相反       前一部分是迴文的 而後一部分能在hashtable 中找到

注意點 在hashtable中查找時 erase掉當前x

先把所有單詞 放入hashtable  對每一個單詞從左到右 遍歷一遍，

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;  
    }  
};