LeetCode 336. Palindrome Pairs

原题链接在这里：https://leetcode.com/problems/palindrome-pairs/

题目：

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"]

题解：

可以合成Palindrome Pairs有几种情况:

1. ["abc", "cba"]

2. ["aabc", "cb"]

3. ["cbaa", "bc"]

要么有个当前string的reverse过来的string也存在，要么当前string的左半部分或者右半部分已经是palindrome, 剩下部分有reverse过来的string存在.

先用HashMap把原有string 和对应index保存。然后对于每一个string拆成left 和 right两个substring, 若是其中一个substring本身就是palindrom, 就看另一个substring的reverse是不是存在.

当然""也是palindrome, 所以如果左右有""存在, 那就是看left, right本身有没有对应的reverse存在.

Note: 要注意["abc", "cba"], 一个substring为""的情况只检查一遍. 不然先检查"abc", left = "", right = "abc", 或者right = "", left = "abc", reverse都存在，就会加[0,1], [1,0]. 等再检查 "cba"时 又会重复加一遍结果. 所以第二个check时要加上right.length() != 0.

注意 i != hm.get(reverseR), 不然会加上[3, 3]. 自己与自己配对的情况.

Time Complexity: O(n * len * len), n = words.length, len时word的平均长度.

Space: O(n), regardless res.

AC Java:

 public class Solution {
     public List<List<Integer>> palindromePairs(String[] words) {
         List<List<Integer>> res = new ArrayList<List<Integer>>();
         if(words == null || words.length < 2){
             return res;
         }

         HashMap<String, Integer> hm = new HashMap<String, Integer>();
         for(int i = 0; i<words.length; i++){
             hm.put(words[i], i);
         }

         for(int i = 0; i<words.length; i++){
             for(int j = 0; j<=words[i].length(); j++){ //j是能到word[i].length()的
                 String left = words[i].substring(0, j);
                 String right = words[i].substring(j);

                 if(isPalindrome(left)){
                     String reverseRight = new StringBuilder(right).reverse().toString();
                     if(hm.containsKey(reverseRight) && hm.get(reverseRight)!=i){
                         List<Integer> item = new ArrayList<Integer>();
                         item.add(hm.get(reverseRight));
                         item.add(i);
                         res.add(item);
                     }
                 }
                 if(isPalindrome(right)){
                     String reverseLeft = new StringBuilder(left).reverse().toString();
                     if(hm.containsKey(reverseLeft) && hm.get(reverseLeft) != i && right.length()!=0){
                         //Addition check is right.length() != 0
                         //Or will add duplicate results, like ["abc", "cba"]
                         List<Integer> item = new ArrayList<Integer>();
                         item.add(i);
                         item.add(hm.get(reverseLeft));
                         res.add(item);
                     }
                 }
             }
         }
         return res;
     }

     private boolean isPalindrome(String s){
         int l = 0;
         int r = s.length()-1;
         while(l<=r){
             if(s.charAt(l++) != s.charAt(r--)){
                 return false;
             }
         }
         return true;
     }
 }

类似Longest Palindromic Substring, Shortest Palindrome.