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"]

解题思路

首先想到的是暴力组合，分别检验所有组合是否为回文，该方法提交后超时，暴力代码如下：

//Time Limit Exceeded
public class Solution {
    public List<List<Integer>> palindromePairs(String[] words) {
        List<List<Integer>> indices = new ArrayList<>();
        for (int i = 0; i < words.length; i++) {
            for (int j = 0; j < words.length; j++) {
                if (j == i) {
                    continue;
                } else {
                    String temp = words[i] + words[j];
                    if (isPalindrome(temp)) {
                        List<Integer> list = new ArrayList<>();
                        list.add(i);
                        list.add(j);
                        indices.add(list);
                    }
                }
            }
        }
        return indices;
    }

    private boolean isPalindrome(String word) {
        for (int i = 0, j = word.length() - 1; i < j; i++, j--) {
            if (word.charAt(i) != word.charAt(j)) {
                return false;
            }
        }
        return true;
    }
}

于是想到另一种思路：

1. 对于空字符串，如果数组中有回文字符串，则将空字符串与回文字符串进行组合。
2. 对于非空字符串，将其进行拆分，如果其中一段为回文，并且数组中可以找到另一段的逆序字符串，则说明可以组合为回文。
   例如：”sssll” = “ss” + “sll”，而且数组中存在”lls”，因此组合”lls”和”sssll”。

注意：组合两个字符串时，需要注意顺序问题。如果字符串前半段为回文，则另一个字符串需要组合在前面；如果字符串后半段为回文，则另一个字符串需要组合在后面。

实现代码

//Runtime: 247 ms
public class Solution {
    public List<List<Integer>> palindromePairs(String[] words) {
        Map<String, Integer> map = new HashMap<>();
        for (int i = 0; i < words.length; i++) {
            map.put(words[i], i);
        }

        List<List<Integer>> indices = new ArrayList<>();
        for (int i = 0; i < words.length; i++) {
            if (words[i].length() == 0) {
                for (Map.Entry<String, Integer> entry : map.entrySet()) {
                    if (isPalindrome(entry.getKey())) {
                        addAll(indices, i, entry.getValue());
                    }
                }
            }

            for (int j = 0; j < words[i].length(); j++) {
                String front = words[i].substring(0, j);
                String back = words[i].substring(j, words[i].length());
                String rfront = reverse(front);
                String rback = reverse(back);
                if (isPalindrome(front) && map.containsKey(rback)) {
                    addAll(indices, map.get(rback), i);
                }
                if (isPalindrome(back) && map.containsKey(rfront)) {
                    addAll(indices, i, map.get(rfront));
                }
            }
        }
        return indices;
    }

    private void addAll(List<List<Integer>> indices, int a, int b) {
        if (a == b) {
            return;
        }
        List<Integer> list = new ArrayList<>();
        list.add(a);
        list.add(b);
        indices.add(list);
    }

    private String reverse(String word) {
        return new StringBuilder(word).reverse().toString();
    }

    private boolean isPalindrome(String word) {
        for (int i = 0, j = word.length() - 1; i < j; i++, j--) {
            if (word.charAt(i) != word.charAt(j)) {
                return false;
            }
        }
        return true;
    }
}