leetcode之Word Pattern

最新推荐文章于 2022-02-24 21:47:19 发布

原创最新推荐文章于 2022-02-24 21:47:19 发布 · 253 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#leetcode #算法 #面试

算法专栏收录该内容

60 篇文章

订阅专栏

本文介绍了一种基于模式匹配的字符串匹配算法，该算法通过建立字符与非空单词之间的双射关系来判断给定的模式是否完全匹配输入字符串。文章提供了一个具体的实现案例，并详细解释了如何使用映射表来确保模式和字符串元素间的对应关系。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

题目：

Given a pattern and a string str, find if str follows the same pattern.

Here follow means a full match, such that there is a bijection between a letter in pattern and a non-empty word in str.

Examples:

pattern = "abba", str = "dog cat cat dog" should return true.
pattern = "abba", str = "dog cat cat fish" should return false.
pattern = "aaaa", str = "dog cat cat dog" should return false.
pattern = "abba", str = "dog dog dog dog" should return false.

Notes:
You may assume pattern contains only lowercase letters, and str contains lowercase letters separated by a single space.

解答：

和上一道题目一样的思路，就是将每对映射到一个数字

class Solution {
public:
    bool wordPattern(string pattern, string str) {
        int lenp = pattern.length();
        int lens = str.length();
        int p = 0;
        int s = 0;
        string tmp = "";
        map<char,int> patternMap;
        map<string,int> strMap;
        while(p < lenp && s < lens)
        {
            while(s < lens && str[s] != ' ')
                tmp += str[s++];
            if(patternMap.find(pattern[p]) == patternMap.end() && strMap.find(tmp) == strMap.end())
                patternMap[pattern[p]] = strMap[tmp] = p + 1;
            else if(patternMap.find(pattern[p]) == patternMap.end() || strMap.find(tmp) == strMap.end())
                return false;
            else
            {
                if(patternMap[pattern[p]] != strMap[tmp])
                    return false;
            }
            p++;
            s++;
            tmp = "";
        }
        if(p != lenp || s != lens + 1)
            return false;
        return true;
     }
};