LeetCode 336. 回文对（哈希map/Trie树）

文章目录

- 1. 题目
- 2. 解题
- - 2.1 哈希map
  - 2.2 Trie树

1. 题目

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"]

来源：力扣（LeetCode）链接：https://leetcode-cn.com/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。

2. 解题

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2.1 哈希map

class Solution {
public:vector<vector<int>> palindromePairs(vector<string>& words) {unordered_map<string, int> w_id;set<int> wdLen;for(int i = 0; i < words.size(); ++i){w_id[words[i]] = i;//字符串idxwdLen.insert(words[i].size());//字符串长度}vector<vector<int>> ans;string front, back, revword;for(int i = 0; i < words.size(); ++i){revword = words[i];//逆序的字符串reverse(revword.begin(),revword.end());if(w_id.count(revword) && w_id[revword] != i)ans.push_back({i, w_id[revword]});//字符串的逆序存在//遍历words[i]可能的子串长度,寻找前部分存在或者后部分存在//且自身剩余的子串为回文int len = words[i].size();for(auto it = wdLen.begin(); *it < len; ++it){front = words[i].substr(0, *it);reverse(front.begin(),front.end());back = words[i].substr(*it);if(w_id.count(front) && ispalind(back))//前缀的逆存在ans.push_back({i, w_id[front]});}for(auto it = wdLen.begin(); *it < len; ++it){front = revword.substr(0, *it);back = revword.substr(*it);if(w_id.count(front) && ispalind(back))//后缀的逆存在ans.push_back({w_id[front], i});}}return ans;}bool ispalind(string& s){int l = 0, r = s.size()-1;while(l < r)if(s[l++] != s[r--])return false;return true;}
};

904 ms 45.6 MB

2.2 Trie树

class trie
{
public:unordered_map<char, trie*> next;int suffix = -1;void insert(string& s, int idx){trie *cur = this;for(int i = s.size()-1; i >= 0; --i)//单词逆序插入{if(!cur->next[s[i]])cur->next[s[i]] = new trie();cur = cur->next[s[i]];}cur->suffix = idx;//结束时记录单词编号}
};
class Solution {
public:vector<vector<int>> palindromePairs(vector<string>& words) {trie * t = new trie(), *cur;vector<vector<int>> ans;string revword;for(int i = 0; i < words.size(); ++i){t->insert(words[i], i);}for(int i = 0; i < words.size(); ++i){int n = words[i].size(), j, k;cur = t;for(j = 0; j < n; ++j){if(cur->suffix != -1 && cur->suffix != i&& ispalind(words[i], j, n-1))//单词的前缀的逆序在trie中，剩余的为回文ans.push_back({i, cur->suffix});if(!cur->next[words[i][j]])break;cur = cur->next[words[i][j]];}for(j = 0; j <= n; ++j)//等号上下只取一次，否则答案有重复的{	// j == n 时包含了完整字符串的情况cur = t;for(k = n-j; k < n; ++k)//遍历单词的后缀{if(!cur->next[words[i][k]])break;cur = cur->next[words[i][k]];}if(k==n && cur->suffix != -1&& cur->suffix != i && ispalind(words[i], 0, n-j-1))//该后缀的逆在trie中，且前部分为回文ans.push_back({cur->suffix, i});}}return ans;}bool ispalind(string s, int l, int r){while(l < r)if(s[l++] != s[r--])return false;return true;}
};