[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 ''