@TOC

前言

给定一组 互不相同 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：words = ["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"]

来源：力扣（LeetCode）
链接：https://leetcode.cn/problems/palindrome-pairs
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

解题思路

原始字符串放入hash表

找前缀

找后缀
所以前缀是回文的情况都试-遍，后缀是回文的情况都试一遍

边界

整体是回文的情况前后拼空字符串

自己的逆序

代码

class Solution {
    public static List<List<Integer>> palindromePairs(String[] words) {
        HashMap<String, Integer> wordset = new HashMap<>();
        for (int i = 0; i < words.length; i++) {
            wordset.put(words[i], i);
        }
        List<List<Integer>> res = new ArrayList<>();
        //{ [6,23] 、 [7,13] }
        for (int i = 0; i < words.length; i++) {
            // i words[i]
            // findAll(字符串，在i位置，wordset) 返回所有生成的结果返回
            res.addAll(findAll(words[i], i, wordset));
        }
        return res;
    }

    public static List<List<Integer>> findAll(String word, int index, HashMap<String, Integer> words) {
        List<List<Integer>> res = new ArrayList<>();
        String reverse = reverse(word);
        Integer rest = words.get("");
        if (rest != null && rest != index && word.equals(reverse)) {
            addRecord(res, rest, index);
            addRecord(res, index, rest);
        }
        int[] rs = manacherss(word);
        int mid = rs.length >> 1;
        for (int i = 1; i < mid; i++) {
            if (i - rs[i] == -1) {
                rest = words.get(reverse.substring(0, mid - i));
                if (rest != null && rest != index) {
                    addRecord(res, rest, index);
                }
            }
        }
        for (int i = mid + 1; i < rs.length; i++) {
            if (i + rs[i] == rs.length) {
                rest = words.get(reverse.substring((mid << 1) - i));
                if (rest != null && rest != index) {
                    addRecord(res, index, rest);
                }
            }
        }
        return res;
    }

    public static void addRecord(List<List<Integer>> res, int left, int right) {
        List<Integer> newr = new ArrayList<>();
        newr.add(left);
        newr.add(right);
        res.add(newr);
    }

    public static int[] manacherss(String word) {
        char[] mchs = manachercs(word);
        int[] rs = new int[mchs.length];
        int center = -1;
        int pr = -1;
        for (int i = 0; i != mchs.length; i++) {
            rs[i] = pr > i ? Math.min(rs[(center << 1) - i], pr - i) : 1;
            while (i + rs[i] < mchs.length && i - rs[i] > -1) {
                if (mchs[i + rs[i]] != mchs[i - rs[i]]) {
                    break;
                }
                rs[i]++;
            }
            if (i + rs[i] > pr) {
                pr = i + rs[i];
                center = i;
            }
        }
        return rs;
    }

    public static char[] manachercs(String word) {
        char[] chs = word.toCharArray();
        char[] mchs = new char[chs.length * 2 + 1];
        int index = 0;
        for (int i = 0; i != mchs.length; i++) {
            mchs[i] = (i & 1) == 0 ? '#' : chs[index++];
        }
        return mchs;
    }

    public static String reverse(String str) {
        char[] chs = str.toCharArray();
        int l = 0;
        int r = chs.length - 1;
        while (l < r) {
            char tmp = chs[l];
            chs[l++] = chs[r];
            chs[r--] = tmp;
        }
        return String.valueOf(chs);
    }
}