leetcode: Substring with Concatenation of All Words

http://oj.leetcode.com/problems/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：

L里每个单词长度一样，写循环就方便了很多。先初始化一个map，统计L里每个单词出现的次数。每次循环如果某个单词没出现或者超出L中出现的错误就中断。

class Solution {
public:
    vector<int> findSubstring(string S, vector<string> &L) {
        int l_size = L.size();
        
        if (l_size <= 0) {
            return vector<int>();
        }
        
        vector<int> result;
        map<string, int> word_count;
        int word_size = L[0].size();
        int i, j;
        
        for (i = 0; i < l_size; ++i) {
            ++word_count[L[i]];
        }
        
        map<string, int> counting;
        
        for (i = 0; i <= (int)S.length() - (l_size * word_size); ++i) {
            counting.clear();
            
            for (j = 0; j < l_size; ++j) {
                string word = S.substr(i + j * word_size, word_size);
                
                if (word_count.find(word) != word_count.end()) {
                    ++counting[word];
                    
                    if (counting[word] > word_count[word]) {
                        break;
                    }
                }
                else {
                    break;
                }
            }
            
            if (j == l_size) {
                result.push_back(i);
            }
        }
        
        return result;
    }
};