336. Palindrome Pairs

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"]

摘自

https://discuss.leetcode.com/topic/40657/150-ms-45-lines-java-solution/2

跟wordladders一样的思想，字串很多，逐个来就是N^2，利用已知的生成再配合hashmap检索能加快速度。

public List<List<Integer>> palindromePairs(String[] words) {    List<List<Integer>> ret = new ArrayList<>();     if (words == null || words.length < 2) return ret;    Map<String, Integer> map = new HashMap<String, Integer>();    for (int i=0; i<words.length; i++) map.put(words[i], i);    for (int i=0; i<words.length; i++) {        // System.out.println(words[i]);        for (int j=0; j<=words[i].length(); j++) { // notice it should be "j <= words[i].length()"            String str1 = words[i].substring(0, j);            String str2 = words[i].substring(j);            if (isPalindrome(str1)) {                String str2rvs = new StringBuilder(str2).reverse().toString();                if (map.containsKey(str2rvs) && map.get(str2rvs) != i) {                    List<Integer> list = new ArrayList<Integer>();                    list.add(map.get(str2rvs));                    list.add(i);                    ret.add(list);                    // System.out.printf("isPal(str1): %s\n", list.toString());                }            }            if (isPalindrome(str2)) {                String str1rvs = new StringBuilder(str1).reverse().toString();                // check "str.length() != 0" to avoid duplicates                if (map.containsKey(str1rvs) && map.get(str1rvs) != i && str2.length()!=0) {                     List<Integer> list = new ArrayList<Integer>();                    list.add(i);                    list.add(map.get(str1rvs));                    ret.add(list);                    // System.out.printf("isPal(str2): %s\n", list.toString());                }            }        }    }    return ret;}private boolean isPalindrome(String str) {    int left = 0;    int right = str.length() - 1;    while (left <= right) {        if (str.charAt(left++) !=  str.charAt(right--)) return false;    }    return true;}

1. The <= in for (int j=0; j<=words[i].length(); j++) is aimed to handle empty string in the input. Consider the test case of ["a", ""];

2. Since we now use <= in for (int j=0; j<=words[i].length(); j++) instead of <. There may be duplicates in the output (consider test case ["abcd", "dcba"]). Therefore I put a str2.length()!=0 to avoid duplicates.