LeetCode--Substring with Concatenation of All Words

通过hash记录L中各个单词出现的次数，由于每个单词仅仅在S的解中出现一次，且单词长度一样，所以还是很好处理的。

但是有一个问题需要注意：

string的length函数返回的size_t是无符号的。参看代码注释1

http://www.cplusplus.com/reference/string/string/length/

class Solution {
public:
    vector<int> findSubstring(string S, vector<string> &L) {
        int lSize = L.size();
        vector<int> res;
        if(lSize == 0)
        {
            return res;
        }

        map<string,int> countL;
        int i;
        for(i = 0 ; i < lSize ; ++i)
        {
            if(countL.find(L[i]) != countL.end())
                countL[L[i]]++;
            else
                countL[L[i]] = 1;
        }
        map<string,int> countS;
        int wordSize = L[0].length();
        //cout<<S.length()<<lSize<<wordSize<<endl;
        int len = S.length();
        //cout<<len-(lSize*wordSize)<<endl;
        for(i = 0 ; i <= len-lSize*wordSize;++i)
//注释1：这里一开始我是直接写的S.length()替代len的，但是当S.length()小于lsize*wordsize的时候，就会出现返回的无符号整型很大很大
        {
            int j;
            countS.clear();
            for(j = 0 ; j < lSize ; ++j)
            {
                string word = S.substr(i+j*wordSize,wordSize);
                if(countS.find(word) != countS.end())
                {
                    ++countS[word];
                    if(countS[word] > countL[word])
                        break;
                }
                else if(countL.find(word) != countL.end())
                {
                    countS[word] = 1;
                }
                else
                    break;
            }
            if(j == lSize)
            {
                res.push_back(i);
            }
        }
        return res;
    }
};