leetcode-hard-array-76. Minimum Window Substring -NO

mycode

不会。。

 

参考：

class Solution(object):
    def minWindow(self, s, t):
        """
        :type s: str
        :type t: str
        :rtype: str
        """
        count1 = collections.defaultdict(int)
        count2 = []
        for char in t:
            count1[char] += 1
            count2.append(char)

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

        for end in range(len(s)):
            if s[end] in count2 :
                #print(end,count1,count2,count)
                count1[s[end]] -= 1
                if count1[s[end]] >= 0:  #<=0时表示出现了多余的t中要求的元素，比如t中两个A，现在出现了第三个A，所以没必要count-1
                    count -= 1
                if count == 0:
                    while True:                      
                        if s[start] in count2  :
                            if count1[s[start]] < 0:  #s中第一个出现t中元素的位置可以被更换啦
                                count1[s[start]] += 1
                            else:  #
                                print(start) 
                                break
                        start += 1
                    if minSize > end - start + 1:
                        minSize = end - start + 1
                        minStart = start
                #print(end,count1,count2,count)
             
        if minSize < len(s) + 1:
            return s[minStart:minStart + minSize]
        else:
            return ''    
        
        class Solution(object):
    def minWindow(self, s, t):
        """
        :type s: str
        :type t: str
        :rtype: str
        """
        count1 = collections.defaultdict(int)
        count2 = []
        for char in t:
            count1[char] += 1
            count2.append(char)

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

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