子串无重复,遍历字符串。设立一个judge子串来判断:
- 如果judge中含有当前字符,更新最大值,更新juge
- 否则judge加上当前字符
class Solution:
def lengthOfLongestSubstring(self, s):
"""
:type s: str
:rtype: int
"""
l1=len(s)
if l1<2:
return l1
judge=''
maxlen=0
for i in range(l1):
if s[i] in judge:
maxlen=max(maxlen,len(judge))
index=judge.find(s[i])
judge=judge[index+1:]+s[i]
else:
judge+=s[i]
return max(maxlen,len(judge))