剑指Offer(Python多种思路实现):最长不含重复字符的子字符串

剑指Offer(Python多种思路实现):最长不含重复字符的子字符串

面试48题:
题目:最长不含重复字符的子字符串

题:请从字符串中找出一个最长的不包含重复字符的子字符串,计算该最长字符串的长度。假设字符串中只包含‘a’-‘z’的字符。例如,在字符串“arabcacfr”中,最长的不含重复字符的子字符串是“acfr”,长度为4。

解题思路一:

class Solution:
    def lengthOfLongestSubstring(self, s):
        """
        :type s: str
        :rtype: int
        """
        start=0
        maxLength=0
        usedChar={}
        for i in range(len(s)):
            if s[i] in usedChar and start<=usedChar[s[i]]:
                start=usedChar[s[i]]+1
            else:
                maxLength=max(maxLength,i-start+1)
            usedChar[s[i]]=i
        
        return maxLength

解题思路二:

def lengthOfLongestSubstring(self, s):
    ans = start = 0
    pos = {}    # last index of element
    for end, c in enumerate(s):
        if c in pos:
            start = max(start, pos[c]+1)
        pos[c] = end
        ans = max(ans, end-start+1)
    return ans
发布了62 篇原创文章 · 获赞 6 · 访问量 3万+

猜你喜欢

转载自blog.csdn.net/weixin_44151089/article/details/104524911
今日推荐