Palindrome Pairs 解答

Question

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"]

Answer

Brute-force的做法是两层for循环，遍历每个string，看有无string可以与之匹配为palindrome。

较好的做法是利用HashMap。

对于String a来说，考虑两种情况：

1. a为空，则a与集合中任意已经对称的string可以组成结果

2. a不为空。

则有三种情况：

a. reverse(a)在集合中

b. a[0..i]对称，且reverse(a[i + 1,...,n])在集合中。如"aaabc", "cba"

c. a[i,..,n]对称，且reverse(a[0,...,i])在集合中。如"abcc", "ba"

public class Solution {
    public List<List<Integer>> palindromePairs(String[] words) {
        List<List<Integer>> result = new ArrayList<>();
        if (words == null || words.length == 0) {
            return result;
        }
        // Record each string position
        Map<String, Integer> map = new HashMap<>();
        Set<Integer> palindromeSet = new HashSet<>();
        int len = words.length;
        for (int i = 0; i < len; i++) {
            map.put(words[i], i);
            if (isPalindrome(words[i])) {
                palindromeSet.add(i);
            }
        }
        // Traverse
        for (int i = 0; i < len; i++) {
            String word = words[i];
            if (word.length() == 0) {
                for (int index : palindromeSet) {
                    addResult(result, i, index);
                }
            }
            int l = word.length();
            for (int j = 0; j < l; j++) {
                String front = word.substring(0, j);
                String back = word.substring(j, l);
                String rFront = reverse(front);
                String rBack = reverse(back);
                if (isPalindrome(front) && map.containsKey(rBack)) {
                    addResult(result, map.get(rBack), i);
                }
                if (isPalindrome(back) && map.containsKey(rFront)) {
                    addResult(result, i, map.get(rFront));
                }
            }
        }
        return result;
    }
    
    private String reverse(String s) {
        StringBuilder sb = new StringBuilder(s);
        return sb.reverse().toString();
    }
    
    private void addResult(List<List<Integer>> result, int start, int end) {
        if (start == end) {
            return;
        }
        List<Integer> indexes = new ArrayList<>();
        indexes.add(start);
        indexes.add(end);
        result.add(indexes);
    }
    
    private boolean isPalindrome(String s) {
        int i = 0;
        int j = s.length() - 1;
        while (i < j) {
            if (s.charAt(i) != s.charAt(j)) {
                return false;
            }
            i++;
            j--;
        }
        return true;
    }
}