leetcode刷题日记——单词规律

[ 题目描述 ]：

[ 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 False

复制代码

      for ch, word in zip(pattern, words):
          if (word in word2ch and word2ch[word] != ch) or (ch in ch2word and ch2word[ch] != word):
              return False
          word2ch[word] = ch
          ch2word[ch] = word
  
      return True