LeetCode刷题实战336：回文对

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于 2021-07-28 13:52:00 发布

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算法的重要性，我就不多说了吧，想去大厂，就必须要经过基础知识和业务逻辑面试+算法面试。所以，为了提高大家的算法能力，这个公众号后续每天带大家做一道算法题，题目就从LeetCode上面选！

今天和大家聊的问题叫做 回文对，我们先来看题面：

https://leetcode-cn.com/problems/palindrome-pairs/

Given a list of unique words, return all the pairs of the distinct indices (i, j) in the given list, so that the concatenation of the two words words[i] + words[j] is a palindrome.

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

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"]

示例 2：

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

示例 3：

输入：words = ["a",""]
输出：[[0,1],[1,0]]

解题

https://cloud.tencent.com/developer/article/1787956

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;
    }
};

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;
    }
};

好了，今天的文章就到这里，如果觉得有所收获，请顺手点个在看或者转发吧，你们的支持是我最大的动力。

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