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leetcode刷题日记——单词规律

news 来源：原创 2025/5/1 5:46:47

[ 题目描述 ]：
在这里插入图片描述
[ 思路 ]：

题目要求判断字符串 s 中的单词是否按照 pattern 这种模式排列
具体思路和 205. 同构字符串基本一致，可以通过 hash 存储来实现
思路二，通过字符串反推 pattern，如果一致，则遵循相同规律，否则不遵循
思路二存在一个问题，pattern 中的字符可能并非按照顺序规律来分配的
例如
代码如下

bool wordPattern(char* pattern, char* s) {int word_count = 0;char* word = strtok(s, " ");char** words = (char**)malloc(strlen(pattern) * sizeof(char*));while (word != NULL) {if(word_count >= strlen(pattern)) return false;words[word_count++] = word;word = strtok(NULL, " ");}char* s_pattern = (char*)malloc(word_count + 1);s_pattern[word_count] = '\0';char current_char = 'a'; char** word_to_char = (char**)malloc(word_count * sizeof(char*));for (int i = 0; i < word_count; i++) {bool found = false;for (int j = 0; j < i; j++) {if (strcmp(words[i], words[j]) == 0) {s_pattern[i] = s_pattern[j];found = true;break;}}if (!found) {s_pattern[i] = current_char++;}}for (int i = 0; i < word_count; i++) {if (s_pattern[i] != pattern[i]) {return false;}}return true;
}

[ 官方题解 ]：

方法一：哈希表；以下对应 Python 3 的代码

class Solution:def wordPattern(self, pattern: str, s: str) -> bool:word2ch = dict()ch2word = dict()words = s.split()if len(pattern) != len(words):return Falsefor ch, word in zip(pattern, words):if (word in word2ch and word2ch[word] != ch) or (ch in ch2word and ch2word[ch] != word):return Falseword2ch[word] = chch2word[ch] = wordreturn True