76. Minimum Window Substring刷题笔记

没做出来，参考的该回答

class Solution:
    def minWindow(self, s: str, t: str) -> str:
        # hash table to store the required char frequency
        need = collections.Counter(t)            

        # total character count we need to care about
        missing = len(t)                         

        # windowStart and windowEnd to be
        windowStart, windowEnd = 0, 0
        i = 0


        # iterate over s starting over index 1
        for j, char in enumerate(s, 1):          
            
            # if char is required then decrease missing
            if need[char] > 0:                   
                missing -= 1

            # decrease the freq of char from need (maybe be negative - which basically denotes
            #   that we have few extra characters which are not required but present in between current window)
            need[char] -= 1                      

            # we found a valid window
            if missing == 0:                     
                # chars from start to find the real windowStart
                while i < j and need[s[i]] < 0:  
                    need[s[i]] += 1
                    i += 1

                # if it's only one char case or curr window is smaller, then update window
                if windowEnd == 0 or j-i < windowEnd-windowStart:  
                    windowStart, windowEnd = i, j

                # now resetting the window to make it invalid
                # sure the first appearing char satisfies need[char]>0
                need[s[i]] += 1          

                # missed this first char, so add missing by 1
                missing += 1                     

                #update i to windowStart+1 for next window
                i += 1                          

        return s[windowStart:windowEnd]
                
                    

运行结果：
速度：48.61%
内存：37.84%