76. Minimum Window Substring

76. Minimum Window Substring

Hard

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).

Example:

Input: S = "ADOBECODEBANC", T = "ABC"
Output: "BANC"

Note:

• If there is no such window in S that covers all characters in T, return the empty string "".
• If there is such window, you are guaranteed that there will always be only one unique minimum window in S.

 

 

笔记：

参考Grandyang

窗口的移动，先向右扩展，然后收缩左边界。c++中array['a']==array[13]可以使用，‘a’代表了a对应的ASCII码，所以在c++中可以使用数组来替代ASCII码，在python中，则不行。array['a']就是dict，其实也就是hash, array['a'] != array[13]。

class Solution(object):
    def minWindow(self, s, t):
        """
        :type s: str
        :type t: str
        :rtype: str
        """
        t_hash = {}
        for v in t:
            if v in t_hash:
                t_hash[v] += 1
            else:
                t_hash[v] = 1
        cnt, left, min_len, res, = 0, 0, float('inf'), ""
        for S_index, S in enumerate(s):
            if S in t_hash:
                t_hash[S] -= 1
                if t_hash[S] >=0 :
                    cnt += 1
            while cnt == len(t):
                if S_index-left+1 < min_len:
                    min_len = S_index - left + 1
                    res = s[left:S_index+1]
                if s[left] in t_hash:
                    t_hash[s[left]] += 1
                    if t_hash[s[left]] > 0:
                        cnt -= 1
                left += 1
        return res