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"]

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

public class Solution {
    public List<List<Integer>> palindromePairs(String[] words) {
        final String EMPTY_STRING = ""; 
        List<List<Integer>> result = new ArrayList<>();
        HashMap<String, Integer> map = new HashMap<>();
        for (int i = 0; i < words.length; i++) {
            map.put(words[i], i);
        }

        for (int i = 0; i < words.length; i++) {
            String s = words[i];

            // if the word is a palindrome, get index of ""
            if (isPalindrome(s)) {
                if (map.containsKey(EMPTY_STRING) && map.get(EMPTY_STRING) != i) {
                    result.add(indexList(i, map.get(EMPTY_STRING)));
                    result.add(indexList(map.get(EMPTY_STRING), i));
                }
            }

            // if the reversed word exists, it is a palindrome, e.g., bat tab
            String reversed = new StringBuilder(s).reverse().toString();
            if (map.containsKey(reversed) && map.get(reversed) != i) {
                result.add(indexList(i, map.get(reversed)));
            }

            for (int k = 1; k < s.length(); k++) {
                String left = s.substring(0, k);
                String right = s.substring(k);

                // if left part is palindrome, find reversed right part
                if (isPalindrome(left)) {
                    String reversedRight = new StringBuilder(right).reverse().toString();
                    if (map.containsKey(reversedRight)) {
                        result.add(indexList(map.get(reversedRight), i));
                    }
                }

                // if right part is a palindrome, find reversed left part
                if (isPalindrome(right)) {
                    String reversedLeft = new StringBuilder(left).reverse().toString();
                    if (map.containsKey(reversedLeft)) {
                        result.add(indexList(i, map.get(reversedLeft)));                        
                    }
                }
            }
        }
        return result;
    }

    public boolean isPalindrome(String s) {
        for (int i = 0, j = s.length() - 1; i < j; i++, j--) {
            if (s.charAt(i) != s.charAt(j)) {
                return false;
            }
        }
        return true;
    }
    
    public List<Integer> indexList(int idx1, int idx2) {
        return Arrays.asList(new Integer[] {idx1, idx2});
    }
}