LeetCode Oj 290. Word Pattern

最新推荐文章于 2017-04-04 11:27:26 发布

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本文介绍了一种基于WordPattern的问题解决方法，通过分析模式与字符串的一致性来判断字符串是否遵循给定的模式。该算法利用哈希映射实现模式与单词间的全匹配对照，并详细解释了其实现过程及关键步骤。

290. Word Pattern

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Total Accepted: 43861
Total Submissions: 144885
Difficulty: Easy

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.

Credits:

class Solution {
public:
    bool wordPattern(string pattern, string str) {
        int vis[500];
        memset(vis , 0, sizeof(vis));
        map<int,string>mp;
        map<string,int>b_mp;
        vector<int>a;
        vector<string>b;
        for(int i=0;i<pattern.length();i++)
           a.push_back(pattern[i]);
        string t="";
        str=str+" ";
        for(int i=0;i<str.length();i++)
           if(str[i]==' ')
            {
                b.push_back(t);
                t="";
            }
           else
            {
                 t=t+str[i];
            }
    if(a.size()!=b.size()) return false;

    for(int i=0;i<a.size();i++)
        {
            if(!vis[a[i]])
            {
                if(b_mp[b[i]]) return false;

                vis[a[i]]=1;
                mp[a[i]]=b[i];
                vis[a[i]]=1;
                b_mp[b[i]]=a[i];
            }
            else
            {
                if(mp[a[i]]!=b[i]) return false;
            }
        }
    return true;
    }
};