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Leetcode03.py
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96 lines (80 loc) · 2.29 KB
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"""
给定一个字符串,请你找出其中不含有重复字符的 最长子串 的长度。
示例 1:
输入: s = "abcabcbb"
输出: 3
解释: 因为无重复字符的最长子串是 "abc",所以其长度为 3。
示例 2:
输入: s = "bbbbb"
输出: 1
解释: 因为无重复字符的最长子串是 "b",所以其长度为 1。
示例 3:
输入: s = "pwwkew"
输出: 3
解释: 因为无重复字符的最长子串是 "wke",所以其长度为 3。
请注意,你的答案必须是 子串 的长度,"pwke" 是一个子序列,不是子串。
示例 4:
输入: s = ""
输出: 0
提示:
0 <= s.length <= 5 * 104
s 由英文字母、数字、符号和空格组成
"""
# 滑动窗口
class Solution:
def lengthOfLongestSubstring(self, s: str) -> int:
if not s:return 0
left = 0
lookup = set()
n = len(s)
max_len = 0
cur_len = 0
for i in range(n):
cur_len += 1
while s[i] in lookup:
lookup.remove(s[left])
left += 1
cur_len -= 1
if cur_len > max_len:max_len = cur_len
lookup.add(s[i])
return max_len
#循环更新
class SolutionTwo:
def lengthOfLongestSubstring(self, s: str) -> int:
ans = ''
tep = ''
for i in s:
if i not in tep:
tep += i
else:
tep = tep[tep.index(i)+1:]
tep +=i
if len(tep)>len(ans):
ans = tep
return len(ans)
class SolutionThree:
def lengthOfLongestSubstring(self, s):
"""
:type s: str
:rtype: int
"""
from collections import defaultdict
lookup = defaultdict(int)
start = 0
end = 0
max_len = 0
counter = 0
while end < len(s):
if lookup[s[end]] > 0:
counter += 1
lookup[s[end]] += 1
end += 1
while counter > 0:
if lookup[s[start]] > 1:
counter -= 1
lookup[s[start]] -= 1
start += 1
max_len = max(max_len, end - start)
return max_len