336. Palindrome Pairs

题目

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:

Input: [“abcd”,“dcba”,“lls”,“s”,“sssll”]
Output: [[0,1],[1,0],[3,2],[2,4]]
Explanation: The palindromes are [“dcbaabcd”,“abcddcba”,“slls”,“llssssll”]

Input: [“bat”,“tab”,“cat”]
Output: [[0,1],[1,0]]
Explanation: The palindromes are [“battab”,“tabbat”]

我的想法

直接brute force对每种组合进行回文串判断复杂度太高了。感觉应该可以用5. Longest Palindromic Substring的suffix tree

解答

leetcode solution: Trie Tree
字典树的典型应用
文字理解、Java实现

class Solution {
    public List<List<Integer>> palindromePairs(String[] words) {
        List<List<Integer>> res = new ArrayList<>();
        if (words == null || words.length == 0) {
            return res;
        }
        TrieNode root = new TrieNode();
        int n = words.length;
        for (int i = 0; i < n; ++i) {
            insert(root, words[i].toCharArray(), i);
        }
        for (int i = 0; i < n; ++i) {
            search(root, words[i].toCharArray(), i, res);
        }
        return res;
    }
    
    private void search(TrieNode root, char[] w, int pos, List<List<Integer>> res) {
        int i1 = 0;
        int n = w.length;
        for (int i = n - 1; i >= 0; --i) {
            if (root.idx >= 0 && root.idx != pos && isPalindrome(w, 0, i)) {
                List<Integer> list = new ArrayList<>();
                list.add(root.idx);
                list.add(pos);
                res.add(list);
            }
            char c = w[i];
            int idx = c - 'a';
            if (root.kids[idx] == null) {
                return;
            } else {
                root = root.kids[idx];
            }
        }
        for (int num : root.toEndIsPalin) {
            if (num == pos) {
                continue;
            }
            List<Integer> list = new ArrayList<>();
            list.add(num);
            list.add(pos);
            res.add(list);
        }
    }
    
    private boolean isPalindrome(char[] c, int s, int e) {
        while (s < e) {
            if (c[s++] != c[e--]) {
                return false;
            }
        }
        return true;
    }
    
    private void insert(TrieNode root, char[] w, int pos) {
        int n = w.length;
        for (int i = 0; i < n; ++i) {
            if (isPalindrome(w, i, n - 1)) {
                root.toEndIsPalin.add(pos);
            }
            int idx = w[i] - 'a';
            if (root.kids[idx] == null) {
                root.kids[idx] = new TrieNode();
            }
            root = root.kids[idx];
        }
        root.toEndIsPalin.add(pos);
        root.idx = pos;
    }
    
    private class TrieNode {
        TrieNode[] kids;
        int idx;
        List<Integer> toEndIsPalin;
        
        TrieNode() {
            idx = -1;
            kids = new TrieNode[26];
            toEndIsPalin = new ArrayList<>();
        }
    }
}