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  • 30. Substring with Concatenation of All Words (String; HashTable)

    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 wordsexactly 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).

    思路: 判断一个值是否包含在一个数组中,首先应该想到将这个数组中的元素放入HashTable,否则每次查找都需要O(n)的时间复杂度。

    时间复杂度:O(n*size),其中n为s的长度,size是指数组words包含多少个元素。当words元素不多的时候,我们可以说时间复杂度是线性的O(n)

    class Solution {
    public:
        vector<int> findSubstring(string s, vector<string>& words) {
            size = words.size();
            sLen = s.length();
            wLen = words[0].length();
            wordsLen = wLen * size;
            for(i = 0; i < size; i++){
                word_counter[words[i]]++;
            }
            
            i = 0;
            while(i+wordsLen<=sLen){
                for(j = 0; j < size; j++){
                    cmpStr = s.substr(i+j*wLen, wLen);
                    if(word_counter.find(cmpStr)==word_counter.end()){ //不在words中,不符合
                        break;
                    }
                    
                    counting[cmpStr]++;
                    if(counting[cmpStr]>word_counter[cmpStr]){ //出现的次数多过words中的次数,不符合
                        break;
                    }
                }
                if(j==size){//找到了一个符合的结果
                    ret.push_back(i);
                }
                counting.clear();
                i++;
            }
            return ret;
        }
    private:
        string cmpStr;
        vector<int> ret;
        map<string,int> word_counter;
        map<string,int> counting;
        int size; //number of words
        int sLen;
        int wLen;
        int wordsLen;
        int i;
        int j;
    };

     

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  • 原文地址:https://www.cnblogs.com/qionglouyuyu/p/4823330.html
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