336. 回文对

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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);
	}
}