Word Pattern - LeetCode

最新推荐文章于 2025-09-07 20:32:33 发布

转载最新推荐文章于 2025-09-07 20:32:33 发布 · 63 阅读

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原文链接：http://www.cnblogs.com/fenshen371/p/4908181.html

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#数据结构与算法

本文介绍了一种基于模式字符串和目标字符串的匹配算法，通过使用两个映射表来判断目标字符串是否完全符合给定的模式。文章提供了具体的代码实现，并通过几个示例验证了算法的有效性。

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.

思路：使用两个map记录是否pattern和str匹配。

注意，map的键值这里不能为0，因为为0时map默认是访问了未创建过的键，因此要从1开始。

此外，for循环里真的可以实现很多东西，用好了会让代码很精简。

 1 class Solution {
 2 public:
 3     bool wordPattern(string pattern, string str) {
 4         map<char, int> dict1;
 5         map<string, int> dict2;
 6         stringstream ss(str);
 7         int i = 0, n = pattern.size();
 8         for (string word; ss >> word; i++)
 9         {
10             if (i == n || dict1[pattern[i]] != dict2[word])
11                 return false;
12             dict1[pattern[i]] = dict2[word] = i + 1;
13         }
14         return i == n;
15     }
16 };