[Leetcode]@python 76. Minimum Window Substring

题目链接

https://leetcode.com/problems/minimum-window-substring/

题目原文

Given a string S and a string T, find the minimum window in S which will contain all the characters in T in complexity O(n).

For example,

S =

"ADOBECODEBANC"

T =

"ABC"

Minimum window is

"BANC"

.

题目大意

给定字符串s和t，要在s中找到一个最小的窗口，其中出现了所有t中的字符

解题思路

t中的字符可能为重复字符，需要先用字典记录每次字符出现的次数；要判断窗口中是否出现了所有字符，首先需要两个指针表示窗口的位置，尾指针不断往后扫，当扫到有一个窗口包含了所有T的字符，然后再收缩头指针，直到不能再收缩为止。最后记录所有可能的情况中窗口最小的。

代码

class Solution(object):
def minWindow(self, s, t):
"""
:type s: str
:type t: str
:rtype: str
"""
count1 = {}
count2 = {}
for char in t:
if char not in count1:
count1[char] = 1
count2[char] = 1
else:
count1[char] += 1
count2[char] += 1

count = len(t)

start = 0
minSize = len(s) + 1
minStart = 0

for end in range(len(s)):
if s[end] in count2 and count2[s[end]] > 0:
count1[s[end]] -= 1
if count1[s[end]] >= 0:
count -= 1
if count == 0:
while True:
if s[start] in count2 and count2[s[start]] > 0:
if count1[s[start]] < 0:
count1[s[start]] += 1
else:
break
start += 1
if minSize > end - start + 1:
minSize = end - start + 1
minStart = start
if minSize < len(s) + 1:
return s[minStart:minStart + minSize]
else:
return ''