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30 Substring with Concatenation of All Words

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30 Substring with Concatenation of All Words

链接:https://leetcode.com/tag/hash-table/
问题描述:
You are given a string, s, and a list of words, words, that are all of the same length. Find all starting indices of substring(s) in s that is a concatenation of each word in words exactly once and without any intervening characters.

For example, given:
s: “barfoothefoobarman”
words: [“foo”, “bar”]

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

Hide Tags Hash Table Two Pointers String

首先明确题目的输入和要求。words里面的单词长度是一样的。words里面的单词会有重复的。要求在s里面,出现所有words里面的单词,不能有多余的一个字符。找到的两个字符串可以有重叠部分。

class Solution {
public:
    vector<int> findSubstring(string s, vector<string>& words)
    {
        vector<int> result;
        int wordlength=words[0].length();
        map<string,int> hm;
        map<string,int> wordmap;
        for(int i=0;i<words.size();i++)
        wordmap[words[i]]++;

        for(int i=0;i<(int)s.length()-(int)words.size()*wordlength+1;i++)
        {
            int j=0;
            for(;j<words.size();j++)
            {
            string word=s.substr(i+j*wordlength,wordlength);
            if(wordmap.find(word)!=wordmap.end())
                hm[word]++;
            else
                break;  
            if(hm[word]>wordmap[word])
                break;
            }
            if(j==words.size())
                result.push_back(i);
            hm.clear();
        }

        return result;
    }
};

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30 Substring with Concatenation of All Words

原文:http://blog.csdn.net/efergrehbtrj/article/details/46815541

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