1 public class Solution { 2 public List<List<Integer>> palindromePairs(String[] words) { 3 List<List<Integer>> ret = new ArrayList<>(); 4 if (words == null || words.length < 2) return ret; 5 Map<String, Integer> map = new HashMap<String, Integer>(); 6 for (int i=0; i<words.length; i++) map.put(words[i], i); 7 for (int i=0; i<words.length; i++) { 8 // System.out.println(words[i]); 9 for (int j=0; j<=words[i].length(); j++) { // notice it should be "j <= words[i].length()" 10 String str1 = words[i].substring(0, j); 11 String str2 = words[i].substring(j); 12 if (isPalindrome(str1)) { 13 String str2rvs = new StringBuilder(str2).reverse().toString(); 14 if (map.containsKey(str2rvs) && map.get(str2rvs) != i) { 15 List<Integer> list = new ArrayList<Integer>(); 16 list.add(map.get(str2rvs)); 17 list.add(i); 18 ret.add(list); 19 // System.out.printf("isPal(str1): %s\n", list.toString()); 20 } 21 } 22 if (isPalindrome(str2)) { 23 String str1rvs = new StringBuilder(str1).reverse().toString(); 24 // check "str.length() != 0" to avoid duplicates 25 if (map.containsKey(str1rvs) && map.get(str1rvs) != i && str2.length()!=0) { 26 List<Integer> list = new ArrayList<Integer>(); 27 list.add(i); 28 list.add(map.get(str1rvs)); 29 ret.add(list); 30 // System.out.printf("isPal(str2): %s\n", list.toString()); 31 } 32 } 33 } 34 } 35 return ret; 36 } 37 38 private boolean isPalindrome(String str) { 39 int left = 0; 40 int right = str.length() - 1; 41 while (left <= right) { 42 if (str.charAt(left++) != str.charAt(right--)) return false; 43 } 44 return true; 45 } 46 }
1. j can reach word[i].length() since it can form an empty string;
2. to aviod duplication ( empty string can cause two found for a pair ,ex [abcd, dbca])
