Monday, November 17, 2014

[LeetCode] Substring with Concatenation of All Words

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).


思路:

和strStr那题的双指针解法类似。关键在于如何判断以任意i起始的S的substring是否整个L的concatenation。这里显然要用到hash table。由于L中可能存在重复的word,所以hash table的key = word,val = count of the word。

在建立好L的hash table后,对每个S[i]进行检查。这里的一个技巧建立一个新的hash table记录已经找到的word。因为L的hash table需要反复利用,不能被修改,并且如果以hash table作为参数进行值传递的化,时间空间消耗都很大。


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class Solution {
public:
    vector<int> findSubstring(string S, vector<string> &L) {
        vector<int> allPos;
        if(L.empty()) return allPos;
        int totalWords = L.size();
        int wordSize = L[0].size();
        int totalLen = wordSize * totalWords;
        if(S.size()<totalLen) return allPos;
        
        unordered_map<string,int> wordCount;
        for(int i=0; i<totalWords; i++)
            wordCount[L[i]]++;
        
        for(int i=0; i<=S.size()-totalLen; i++) {
            if(checkSubstring(S, i, wordCount, wordSize, totalWords))
                allPos.push_back(i);
        }
        return allPos;
    }
    
    bool checkSubstring(string S, int start, unordered_map<string,int> &wordCount, int wordSize, int totalWords) {
        if(S.size()-start+1 < wordSize*totalWords) return false;
        unordered_map<string,int> wordFound;
        
        for(int i=0; i<totalWords; i++) {
            string curWord = S.substr(start+i*wordSize,wordSize);
            if(!wordCount.count(curWord)) return false;
            wordFound[curWord]++;
            if(wordFound[curWord]>wordCount[curWord]) return false;
        }
        return true;
    }
};

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