LeetCode 290 Word Pattern

Posted 2020-08-26 SillyVicky

Problem:

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:

1. pattern = "abba", str = "dog cat cat dog" should return true.
2. pattern = "abba", str = "dog cat cat fish" should return false.
3. pattern = "aaaa", str = "dog cat cat dog" should return false.
4. 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.

Summary:

判断所给字符串str中单词的形式与pattern中是否相符。

Solution:

以key为pattern中的char，value为str中的单词构成Hash Table，每加进一个新单词，判断原来若存在相同key的单词是否与新单词相同。

最终需排除不同key，value相同的情况。

 1 class Solution {
 2 public:
 3     bool wordPattern(string pattern, string str) {
 4         vector<string> s;
 5         unordered_map<char, string> m;
 6         
 7         int flag = 0;
 8         for (int i = 0; i < str.size(); i++) {
 9             if (str[i] == ‘ ‘) {
10                 s.push_back(str.substr(flag, i - flag));
11                 flag = i + 1;
12             }
13         }
14         s.push_back(str.substr(flag, str.size() - flag));
15         
16         if (pattern.size() != s.size()) {
17             return false;
18         }
19         
20         for (int i = 0; i < pattern.size(); i++) {
21             if (m[pattern[i]] == "") {
22                 m[pattern[i]] = s[i];
23             }
24             else if (m[pattern[i]] != s[i]){
25                 return false;
26             }
27         }
28         
29         for (unordered_map<char, string>::iterator i = m.begin(); i != m.end(); i++) {
30             for (unordered_map<char, string>::iterator j = m.begin(); j != m.end(); j++) {
31                 if (i->second == j->second && i->first != j->first) {
32                     return false;
33                 }
34             }
35         }
36         
37         return true;
38     }
39 };