LeetCode 336. 回文对（哈希map/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

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idx
    		wdLen.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;
    }
};

940 ms 141.3 MB

trie 改用数组 trie* next[26] = {NULL}; 提高运行效率

280 ms 208.5 MB

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