JAVA程序设计：回文对（LeetCode：336）

给定一组唯一的单词， 找出所有不同 的索引对(i, j)，使得列表中的两个单词， words[i] + words[j] ，可拼接成回文串。

示例 1:

输入: ["abcd","dcba","lls","s","sssll"]
输出: [[0,1],[1,0],[3,2],[2,4]] 
解释: 可拼接成的回文串为 ["dcbaabcd","abcddcba","slls","llssssll"]
示例 2:

输入: ["bat","tab","cat"]
输出: [[0,1],[1,0]] 
解释: 可拼接成的回文串为 ["battab","tabbat"]

思路：我是纯字典树过的，再次刷这道题时参考了官方题解，通过马拉车优化回文串查找。

class Solution {

    class trie {
        class node {
            int[] c = new int[26];
            int flag;

            public node() {
                flag = -1;
            }
        }

        List tree = new ArrayList<>();

        public trie() {
            tree.add(new node());
        }

        public void insert(String s, int id) {
            int len = s.length(), add = 0;
            for (int i = 0; i < len; i++) {
                int x = s.charAt(i) - 'a';
                if (tree.get(add).c[x] == 0) {
                    tree.add(new node());
                    tree.get(add).c[x] = tree.size() - 1;
                }
                add = tree.get(add).c[x];
            }
            tree.get(add).flag = id;
        }

        public int[] query(String s) {
            int len = s.length(), add = 0;
            int[] ret = new int[len + 1];
            Arrays.fill(ret, -1);
            for (int i = 0; i < len; i++) {
                ret[i] = tree.get(add).flag;
                int x = s.charAt(i) - 'a';
                if (tree.get(add).c[x] == 0)
                    return ret;
                add = tree.get(add).c[x];
            }
            ret[len] = tree.get(add).flag;
            return ret;
        }
    }

    public List> palindromePairs(String[] words) {

        trie trie1 = new trie();
        trie trie2 = new trie();

        int n = words.length;
        for (int i = 0; i < n; i++) {
            trie1.insert(words[i], i);
            StringBuilder str = new StringBuilder(words[i]);
            str.reverse();
            trie2.insert(str.toString(), i);
        }

        List> ret = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            int[][] rec = manacher(words[i]);

            int[] id1 = trie2.query(words[i]);
            words[i] = new StringBuilder(words[i]).reverse().toString();
            int[] id2 = trie1.query(words[i]);

            int m = words[i].length();

            int allid = id1[m];
            if (allid != -1 && allid != i)
                ret.add(Arrays.asList(i, allid));
            for (int j = 0; j < m; j++) {
                if (rec[j][0] != 0) {
                    int leftid = id2[m - j - 1];
                    if (leftid != -1 && leftid != i)
                        ret.add(Arrays.asList(leftid, i));
                }
                if (rec[j][1] != 0) {
                    int rightid = id1[j];
                    if (rightid != -1 && rightid != i)
                        ret.add(Arrays.asList(i, rightid));
                }
            }
        }

        return ret;
    }

    public int[][] manacher(String s) {
        int n = s.length();
        StringBuilder tmp = new StringBuilder("#");
        for (int i = 0; i < n; i++) {
            if (i > 0) tmp.append('*');
            tmp.append(s.charAt(i));
        }
        tmp.append('!');
        int m = n * 2;
        int[] len = new int[m];
        int[][] ret = new int[n][2];
        int p = 0, maxn = -1;
        for (int i = 1; i < m; i++) {
            len[i] = maxn >= i ? Math.min(len[2 * p - i], maxn - i) : 0;
            while (tmp.charAt(i - len[i] - 1) == tmp.charAt(i + len[i] + 1))
                len[i]++;
            if (i + len[i] > maxn) {
                p = i;
                maxn = i + len[i];
            }
            if (i - len[i] == 1) ret[(i + len[i]) / 2][0] = 1;
            if (i + len[i] == m - 1) ret[(i - len[i]) / 2][1] = 1;
        }
        return ret;
    }
}