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[LeetCode] Minimum Window Substring — python

2017年01月10日 编程语言 ⁄ 共 1186字 ⁄ 字号 评论关闭
程序员最喜欢的衬衫

最近在用python刷leetcode的题,发现写起来比c++快多了~

这道题应该是双指针s1,s2,从头扫描,每一步先移动s2,使s1-s2范围内包含答案,随后移动s1,使包含结果的字符串最短,

刚开始做的时候TLE了,因为check的时候把dict过了一遍,太慢了,后来改为移动s1的时候,只check删除的s1指向的字符是否还够。

代码如下:

class Solution:
    # @return a string
    def minWindow(self, S, T):
        if not T or not S or len(T)>len(S):
            return ''
        res=''
        s1=0
        wordS={}
        wordT={}
        for i in T:
            self.addToDict(i,wordT)
        for s2 in xrange(len(S)):
            self.addToDict(S[s2],wordS)
            if not self.check(wordT,wordS):
                break
            if s2==len(S)-1:
                return res
        s2+=1
        while s2<=len(S):
            while s1<s2 and self.checkOne(wordT,wordS,S[s1])>0:
                #contain all
                # print s1,s2,S[s1],wordS
                wordS[S[s1]]-=1
                s1+=1
            print s1,s2,wordS
            if (not res) or len(S[s1:s2])<len(res):
                res=S[s1:s2]

            if s2>=len(S):
                break
            self.addToDict(S[s2],wordS)
            s2+=1
                
            
            
        ## TLE
        # for s1 in xrange(len(S)):
        #     while s2<len(S) and self.check(wordT,wordS):
        #         s2+=1
        #         self.addToDict(S[s2-1],wordS)
        #     if not self.check(wordT,wordS):#contain all
        #         if (not res) or len(S[s1:s2+1])<len(res):
        #             res=S[s1:s2+1]
        #     s1+=1
        #     wordS[S[s1-1]]-=1
        return res

    def check(self, dictT, dict2):
        for key in dictT:
            try:
                if dictT[key]>dict2[key]:
                    return key 
            except Exception, e:
                return key
        return ''

    def checkOne(self, dictT, dict2, key):
        if key in dictT:
            try:
                return dict2[key]-dictT[key]
            except Exception, e:
                return -1
        else:
            return 1
          
    def addToDict(self, item, dict):
        dict.setdefault(item,0)
        dict[item]+=1

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