Algorithm to Check if A String Matches a Pattern
- 时间:2020-10-09 18:35:39
- 分类:网络文摘
- 阅读:85 次
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: trueExample 2:
Input:pattern = “abba”, str = “dog cat cat fish”
Output: falseExample 3:
Input: pattern = “aaaa”, str = “dog cat cat dog”
Output: falseExample 4:
Input: pattern = “abba”, str = “dog dog dog dog”
Output: falseNotes:
You may assume pattern contains only lowercase letters, and str contains lowercase letters that may be separated by a single space.
Check Pattern Using Hash Maps
We can use two hash maps to associate with pattern to words and vice versa. Spliting the string by delimiter space, and go through each token/word to see if it matches the existing character. If not, we store the new mapping. If there is a mapping already – we check if it is the same and return false immediately when mapping does not match.
Some edge cases are to be handled – checking if the pattern sizes same as the number of the words in the sentence. Also, the same words cannot be mapped to different token/pattern. And also, different words cannot be mapped to a same token/pattern.
The following C++ uses istringstream to process a string into words/tokens. The complexity is O(N) where N is the number of the characters in the string – and O(N) space as we are using two hash maps.
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 | class Solution { public: bool wordPattern(string pattern, string str) { istringstream ss(str); string word; unordered_map<char, string> mapping; unordered_map<string, char> words; int i = 0; while (ss >> word) { if (i >= pattern.size()) { return false; } char pat = pattern[i ++]; if (mapping.find(pat) != mapping.end()) { if (word != mapping[pat]) { return false; } } if ((words.find(word) != words.end()) && (words[word] != pat)) { return false; } words[word] = pat; mapping[pat] = word; } return i == pattern.size(); } }; |
class Solution { public: bool wordPattern(string pattern, string str) { istringstream ss(str); string word; unordered_map<char, string> mapping; unordered_map<string, char> words; int i = 0; while (ss >> word) { if (i >= pattern.size()) { return false; } char pat = pattern[i ++]; if (mapping.find(pat) != mapping.end()) { if (word != mapping[pat]) { return false; } } if ((words.find(word) != words.end()) && (words[word] != pat)) { return false; } words[word] = pat; mapping[pat] = word; } return i == pattern.size(); } };
The following is the Python implementation of the word pattern algorithm – the code is concise but only using one dictionary/hashmap:
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 | class Solution: def wordPattern(self, pattern: str, str: str) -> bool: arr = str.split(' ') if (len(arr) != len(pattern)): return False data = {} idx = 0 for s in pattern: if not s in data: if arr[idx] in data.values(): return False data[s] = arr[idx] else: if data[s] != arr[idx]: return False idx += 1 return True |
class Solution: def wordPattern(self, pattern: str, str: str) -> bool: arr = str.split(' ') if (len(arr) != len(pattern)): return False data = {} idx = 0 for s in pattern: if not s in data: if arr[idx] in data.values(): return False data[s] = arr[idx] else: if data[s] != arr[idx]: return False idx += 1 return True
The algorithm complexity is O(N^2) as we are using in operator to check if a word exists in an array.
–EOF (The Ultimate Computing & Technology Blog) —
推荐阅读:音乐数3、4、6 试管中的所有空间被细胞占满 走马灯数 求往返的平均速度 SEO网站做关键词上首页需要多长时间? 手机网站在建设中 能够提升利用率的一些要点 做什么网站赚钱?游戏类网站可以考虑 官方回应国内版n号房调查:严厉追究法律责任! 分付,不是微信版“花呗”! 可往湖北寄快递了!湖北快递全面恢复!
- 评论列表
-
- 添加评论