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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"]
分析：http://bookshadow.com/weblog/2016/03/10/leetcode-palindrome-pairs/
利用字典wmap保存单词 -> 下标的键值对
遍历单词列表words，记当前单词为word，下标为idx：
1). 若当前单词word本身为回文，且words中存在空串，则将空串下标bidx与idx加入答案
2). 若当前单词的逆序串在words中，则将逆序串下标ridx与idx加入答案
3). 将当前单词word拆分为左右两半left，right。
3.1) 若left为回文，并且right的逆序串在words中，则将right的逆序串下标rridx与idx加入答案
3.2) 若right为回文，并且left的逆序串在words中，则将left的逆序串下标idx与rlidx加入答案
1 public class Solution { 2 public List<List<Integer>> palindromePairs(String[] words) { 3 List<List<Integer>> result = new ArrayList<List<Integer>>(); 4 5 HashMap<String, Integer> map = new HashMap<String, Integer>(); 6 for (int i = 0; i < words.length; i++) { 7 map.put(words[i], i); 8 } 9 10 for (int i = 0; i < words.length; i++) { 11 String s = words[i]; 12 13 // if the word is a palindrome, get index of "" 14 if (isPalindrome(s)) { 15 if (map.containsKey("") && map.get("") != i) { 16 ArrayList<Integer> l = new ArrayList<Integer>(); 17 l.add(i); 18 l.add(map.get("")); 19 result.add(l); 20 21 l = new ArrayList<Integer>(); 22 23 l.add(map.get("")); 24 l.add(i); 25 result.add(l); 26 } 27 } 28 29 // if the reversed word exists, it is a palindrome, e.g., bat tab 30 String reversed = new StringBuilder(s).reverse().toString(); 31 if (map.containsKey(reversed) && map.get(reversed) != i) { 32 ArrayList<Integer> l = new ArrayList<Integer>(); 33 l.add(i); 34 l.add(map.get(reversed)); 35 result.add(l); 36 } 37 38 for (int k = 1; k < s.length(); k++) { 39 String left = s.substring(0, k); 40 String right = s.substring(k); 41 42 // if left part is palindrome, find reversed right part 43 if (isPalindrome(left)) { 44 String reversedRight = new StringBuilder(right).reverse().toString(); 45 if (map.containsKey(reversedRight)) { 46 ArrayList<Integer> l = new ArrayList<Integer>(); 47 l.add(map.get(reversedRight)); 48 l.add(i); 49 result.add(l); 50 } 51 } 52 53 // if right part is a palindrome, find reversed left part 54 if (isPalindrome(right)) { 55 String reversedLeft = new StringBuilder(left).reverse().toString(); 56 if (map.containsKey(reversedLeft)) { 57 ArrayList<Integer> l = new ArrayList<Integer>(); 58 l.add(i); 59 l.add(map.get(reversedLeft)); 60 result.add(l); 61 } 62 } 63 } 64 } 65 return result; 66 } 67 68 public boolean isPalindrome(String s) { 69 70 int i = 0; 71 int j = s.length() - 1; 72 73 while (i < j) { 74 if (s.charAt(i) != s.charAt(j)) { 75 return false; 76 } 77 i++; 78 j--; 79 } 80 return true; 81 } 82 }
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原文地址：http://www.cnblogs.com/beiyeqingteng/p/5724408.html
