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  • [leetcode]Substring with Concatenation of All Words @ Python

    原题地址:https://oj.leetcode.com/problems/substring-with-concatenation-of-all-words/

    题意:

    You are given a string, S, and a list of words, L, that are all of the same length. Find all starting indices of substring(s) in S that is a concatenation of each word in L exactly once and without any intervening characters.

    For example, given:
    S"barfoothefoobarman"
    L["foo", "bar"]

    You should return the indices: [0,9].
    (order does not matter).

    解题思路:使用一个字典统计一下L中每个单词的数量。由于每个单词的长度一样,以题中给的例子而言,可以3个字母3个字母的检查,如果不在字典中,则break出循环。有一个技巧是建立一个临时字典curr,用来统计S中那些在L中的单词的数量,必须和L中单词的数量相等,否则同样break。

    代码:

    class Solution:
        # @param S, a string
        # @param L, a list of string
        # @return a list of integer
        def findSubstring(self, S, L):
            words={}
            wordNum=len(L)
            for i in L:
                if i not in words:
                    words[i]=1
                else:
                    words[i]+=1
            wordLen=len(L[0])
            res=[]
            for i in range(len(S)+1-wordLen*wordNum):
                curr={}; j=0
                while j<wordNum:
                    word=S[i+j*wordLen:i+j*wordLen+wordLen]
                    if word not in words: 
                        break
                    if word not in curr: 
                        curr[word]=1
                    else:
                        curr[word]+=1
                    if curr[word]>words[word]: break
                    j+=1
                if j==wordNum: res.append(i)
            return res
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  • 原文地址:https://www.cnblogs.com/zuoyuan/p/3779978.html
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