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"

1 public class Solution { 2     public List
     
      
       > palindromePairs(String[] words) { 3         List
       
        
         > result = new ArrayList<>(); 4         if (words == null || words.length == 0) { 5             return result; 6         } 7         // Record each string position 8         Map
         
           map = new HashMap<>(); 9         Set
          
            palindromeSet = new HashSet<>();10 int len = words.length;11 for (int i = 0; i < len; i++) {12 map.put(words[i], i);13 if (isPalindrome(words[i])) {14 palindromeSet.add(i);15 }16 }17 // Traverse18 for (int i = 0; i < len; i++) {19 String word = words[i];20 if (word.length() == 0) {21 for (int index : palindromeSet) {22 addResult(result, i, index);23 }24 }25 int l = word.length();26 for (int j = 0; j < l; j++) {27 String front = word.substring(0, j);28 String back = word.substring(j, l);29 String rFront = reverse(front);30 String rBack = reverse(back);31 if (isPalindrome(front) && map.containsKey(rBack)) {32 addResult(result, map.get(rBack), i);33 }34 if (isPalindrome(back) && map.containsKey(rFront)) {35 addResult(result, i, map.get(rFront));36 }37 }38 }39 return result;40 }41 42 private String reverse(String s) {43 StringBuilder sb = new StringBuilder(s);44 return sb.reverse().toString();45 }46 47 private void addResult(List
           
            
             > result, int start, int end) {48 if (start == end) {49 return;50 }51 List
             
               indexes = new ArrayList<>();52 indexes.add(start);53 indexes.add(end);54 result.add(indexes);55 }56 57 private boolean isPalindrome(String s) {58 int i = 0;59 int j = s.length() - 1;60 while (i < j) {61 if (s.charAt(i) != s.charAt(j)) {62 return false;63 }64 i++;65 j--;66 }67 return true;68 }69 }