792. 匹配子序列的单词数

暴力算法果断超时

class Solution(object):
    def isSubsequence(self, s, t):
        """
        :type s: str
        :type t: str
        :rtype: bool
        """
        l1 = len(s)
        l2 = len(t)
        p1 = 0
        p2 = 0
        while p1< l1 and p2<l2:
            if s[p1] == t[p2]:
                p1+=1
                p2+=1
            else:
                p2+=1
                
        if p1 == l1:
            return True
        else:
            return False
    def numMatchingSubseq(self, S, words):
        """
        :type S: str
        :type words: List[str]
        :rtype: int
        """
        sum = 0
        for word in words:
            if self.isSubsequence(word,S) == True:
                sum+=1
        return sum
        
    原文作者:cptn3m0
    原文地址: https://www.jianshu.com/p/6af2277fe2e9
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