[LeetCode] 809. Expressive Words_Medium tag: array, two pointers

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Sometimes people repeat letters to represent extra feeling. For example:

• "hello" -> "heeellooo"
• "hi" -> "hiiii"

In these strings like "heeellooo", we have groups of adjacent letters that are all the same: "h", "eee", "ll", "ooo".

You are given a string s and an array of query strings words. A query word is stretchy if it can be made to be equal to s by any number of applications of the following extension operation: choose a group consisting of characters c, and add some number of characters c to the group so that the size of the group is three or more.

• For example, starting with "hello", we could do an extension on the group "o" to get "hellooo", but we cannot get "helloo" since the group "oo" has a size less than three. Also, we could do another extension like "ll" -> "lllll" to get "helllllooo". If s = "helllllooo", then the query word "hello" would be stretchy because of these two extension operations: query = "hello" -> "hellooo" -> "helllllooo" = s.

Return the number of query strings that are stretchy.

Example 1:

Input: s = "heeellooo", words = ["hello", "hi", "helo"]
Output: 1
Explanation: 
We can extend "e" and "o" in the word "hello" to get "heeellooo".
We can't extend "helo" to get "heeellooo" because the group "ll" is not size 3 or more.

Example 2:

Input: s = "zzzzzyyyyy", words = ["zzyy","zy","zyy"]
Output: 3

Constraints:

• 1 <= s.length, words.length <= 100
• 1 <= words[i].length <= 100
• s and words[i] consist of lowercase letters.

Ideas:

1. helper function, check s and word 

2. 得到word[start_word]和s[start_s], 如果相等,再看他们的次数是否符合题意

3. 最后得到sum 并返回

Code:

class Solution:
    def expressiveWords(self, s: str, words: List[str]) -> int:
        return sum([self.exWord(s, 0, word, 0) for word in words])
    
    
    def exWord(self, s, start_s,  word, start_word) -> bool:
        l_s, l_word = len(s), len(word)
        if start_s == l_s and start_word == l_word: return True
        elif start_s >= l_s or start_word >= l_word:
            return False
        char = s[start_s]
        if char == word[start_word]:
            count1, count2 = 0, 0
            while start_s < l_s and s[start_s] == char:
                count1 += 1
                start_s += 1
            while start_word < l_word and word[start_word] == char:
                count2 += 1
                start_word += 1
            if count1 == count2 or (count1 >= 3 and count1 > count2):
                return self.exWord(s, start_s, word, start_word)
        return False
        
        

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