Palindrome Pairs 解答

Question

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"]

Answer

Brute-force的做法是两层for循环，遍历每个string，看有无string可以与之匹配为palindrome。

较好的做法是利用HashMap。

对于String a来说，考虑两种情况：

1. a为空，则a与集合中任意已经对称的string可以组成结果

2. a不为空。

则有三种情况：

a. reverse(a)在集合中

b. a[0..i]对称，且reverse(a[i + 1,...,n])在集合中。如"aaabc", "cba"

c. a[i,..,n]对称，且reverse(a[0,...,i])在集合中。如"abcc", "ba"

 public class Solution {
     public List<List<Integer>> palindromePairs(String[] words) {
         List<List<Integer>> result = new ArrayList<>();
         if (words == null || words.length == 0) {
             return result;
         }
         // Record each string position
         Map<String, Integer> map = new HashMap<>();
         Set<Integer> palindromeSet = new HashSet<>();
         int len = words.length;
         for (int i = 0; i < len; i++) {
             map.put(words[i], i);
             if (isPalindrome(words[i])) {
                 palindromeSet.add(i);
             }
         }
         // Traverse
         for (int i = 0; i < len; i++) {
             String word = words[i];
             if (word.length() == 0) {
                 for (int index : palindromeSet) {
                     addResult(result, i, index);
                 }
             }
             int l = word.length();
             for (int j = 0; j < l; j++) {
                 String front = word.substring(0, j);
                 String back = word.substring(j, l);
                 String rFront = reverse(front);
                 String rBack = reverse(back);
                 if (isPalindrome(front) && map.containsKey(rBack)) {
                     addResult(result, map.get(rBack), i);
                 }
                 if (isPalindrome(back) && map.containsKey(rFront)) {
                     addResult(result, i, map.get(rFront));
                 }
             }
         }
         return result;
     }

     private String reverse(String s) {
         StringBuilder sb = new StringBuilder(s);
         return sb.reverse().toString();
     }

     private void addResult(List<List<Integer>> result, int start, int end) {
         if (start == end) {
             return;
         }
         List<Integer> indexes = new ArrayList<>();
         indexes.add(start);
         indexes.add(end);
         result.add(indexes);
     }

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