题目

输入: s1 = “ab” s2 = “eidbaooo”
输出: True
解释: s2 包含 s1 的排列之一 (“ba”).

题解

1 sort字符串对比

class Solution:
    def checkInclusion(self, s1: str, s2: str) -> bool:
        # 窗口为len(s1)，然后对窗口内的以及s1都sort后对比

        len1 = len(s1)
        pattern = "".join(sorted(s1))
        for i in range(len(s2)):
            tmp = s2[i:len1+i]
            if "".join(sorted(tmp)) == pattern:
                return True
        return False

2 字典统计窗口字母 key = 字母，value = 出现次数

• 因为是小写字母,所以可以映射到0-25

class Solution:
    def checkInclusion(self, s1: str, s2: str) -> bool:
        if len(s2) < len(s1):
            return False
        # ord('a') = 97
        # 计算s1的字符状态
        stat1 = [0] * 26
        for ch in s1:
            stat1[ord(ch) - 97] += 1
        # 初始化滑动窗口
        stat2 = [0] * 26
        for i in range(len(s1)):
            stat2[ord(s2[i]) - 97] += 1
        if stat1 == stat2:
            return True
        # 移动滑动窗口
        for i in range(len(s1), len(s2)):
            stat2[ord(s2[i]) - 97] += 1
            stat2[ord(s2[i - len(s1)]) - 97] -= 1
            if stat1 == stat2:
                return True
        return False

            

标签：return,stat2,Offer,s2,s1,len,II,014,ord
来源： https://blog.csdn.net/weixin_43702920/article/details/120289436