LeetCode #290 - Word Pattern

最新推荐文章于 2021-02-26 18:39:59 发布

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本文介绍了一种判断字符串是否遵循给定模式的方法。通过建立字符与单词间的映射关系，确保模式与字符串间存在一一对应的关系。举例说明了如何使用此方法解决不同情况下的模式匹配问题。

题目描述：

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.

Example 1:

Input: pattern = "abba", str = "dog cat cat dog"

Output: true

Example 2:

Input:pattern = "abba", str = "dog cat cat fish"

Output: false

Example 3:

Input: pattern = "aaaa", str = "dog cat cat dog"

Output: false

Example 4:

Input: pattern = "abba", str = "dog dog dog dog"

Output: false

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

字符与单词的模式匹配，和单词之间的模式匹配是一样的方法。

class Solution {
public:
    bool wordPattern(string pattern, string str) {
        vector<string> v=split(str);
        unordered_map<string,int> hash1;
        unordered_map<char,int> hash2;
        if(pattern.size()!=v.size()) return false;
        for(int i=0;i<v.size();i++)
        {
            if(hash1.count(v[i])!=hash2.count(pattern[i])) return false;
            else if(hash1.count(v[i])>0&&hash1[v[i]]!=hash2[pattern[i]]) return false;
            hash1[v[i]]=i;
            hash2[pattern[i]]=i;
        }
        return true;
    }
    
    vector<string> split(string s)
    {
        vector<string> result;
        int i=0;
        int j=0;
        while(j<s.size())
        {
            if(s[j]==' ') 
            {
                result.push_back(s.substr(i,j-i));
                j++;
                i=j;
            }
            else if(j==s.size()-1) 
            {
                result.push_back(s.substr(i,j-i+1));
                break;
            }
            else j++;
        }
        return result;
    }
};