# 模板

left,right = 0,0
while right < len(s):
windows.append(s[right])
right += 1
while (windows needs shrink):
window.pop(0)
left += 1


# leetcode 76 最小覆盖子串

class Solution:
def minWindow(self, s: str, t: str) -> str:
from collections import Counter
needs = Counter(t)
windows = {i:0 for i in t}
left, right = 0, 0
valid = 0
# 记录最小覆盖子串的起始位置以及长度
start = 0
leng = float("inf")
while right < len(s):
c = s[right]
right += 1
if c in needs:
windows[c] += 1
if needs[c] == windows[c]:
valid += 1
while valid == len(needs):
if right - left < leng:
start = left
leng = right - left
d = s[left]
left += 1
if d in needs:
if windows[d] == needs[d]:
valid -= 1
windows[d] -= 1
return "" if leng == float("inf") else s[start:start+leng]


# leetcode 567 字符串排列

class Solution:
def checkInclusion(self, s1: str, s2: str) -> bool:
from collections import Counter
needs = Counter(s1)
windows = {i:0 for i in s1}
left, right = 0, 0
valid = 0
while right < len(s2):
c = s2[right]
right += 1
if c in needs:
windows[c] += 1
if needs[c] == windows[c]:
valid += 1
while right - left > len(s1) - 1: # 窗口中只能有len(s1)个元素，
if valid == len(needs):
print(s2[left:right])
return True
d = s2[left]
left += 1
if d in needs:
if windows[d] == needs[d]:
valid -= 1
windows[d] -= 1
return False


# leetcode 438 找所有字母异位词

class Solution:
def findAnagrams(self, s: str, p: str) -> List[int]:
from collections import Counter
needs = Counter(p)
windows = {i:0 for i in p}
left, right = 0, 0
valid = 0
res = []
while right < len(s):
c = s[right]
right += 1
if c in needs:
windows[c] += 1
if needs[c] == windows[c]:
valid += 1
while right - left > len(p) - 1: # 窗口中只能有两个元素，
if valid == len(needs):
print(s[left:right])
res.append(left)
d = s[left]
left += 1
if d in needs:
if windows[d] == needs[d]:
valid -= 1
windows[d] -= 1
return res


# leetcode 3 最长无重复子串

class Solution:
def lengthOfLongestSubstring(self, s: str) -> int:
if len(s) == 0:
return 0
if len(s) == 1:
return 1
from collections import defaultdict
windows = defaultdict(int)
left, right = 0,0
res = float("-inf")
while right < len(s):
c = s[right]
right += 1
windows[c] += 1
while windows[c] > 1:
d = s[left]
left += 1
windows[d] -= 1
res = max(res, right - left)
return res

posted @ 2021-02-04 10:03  西西嘛呦  阅读(452)  评论(0编辑  收藏  举报