【leetcode】最小覆盖子串

时间复杂度还是高，双指针有待优化

class Solution:
    def minWindow(self, s: str, t: str) -> str:
        se = set(list(t))                
        p = []
        ans = ""
        maxn = len(s) + 1
        from collections import defaultdict
        last = defaultdict(list)
        vis = defaultdict(int)
        raw = defaultdict(int)
        for i in range(len(t)):
            raw[t[i]] += 1
        for i in range(len(s)):            
            if(s[i] in se):                
                if(vis[s[i]] < raw[s[i]]):
                    vis[s[i]] += 1
                else:
                    cur = last[s[i]][0]
                    p.remove(cur)   
                    last[s[i]].remove(cur)   
                last[s[i]].append(i)
                p.append(i)                   
            if(len(p) == len(t)):                
                st = p[0]
                ed = p[len(p)-1]
                if(ed - st + 1 < maxn):
                    maxn = ed - st + 1
                    ans = s[st:ed + 1]           
        return ans