substring-with-concatenation-of-all-words

【题目描述】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).
给你一个字符串，和一个字符串的数组，需要返回一个该字符串的索引组成的数组，返回的索引有如下性质：从每个索引开始，长度为L的字串需要精确包含字符串数组中的所有字符串（不多不少）。L 为字符串数组中所有字符串长度之和。

【解题思路】使用一个map，键为字符串数组中的字符串，值为该字符串在字符串数组中出现的次数。遍历字符串s，寻找和字符串数组中的字符串相同的字串，找到后map中的值减一，否则重新初始化map，从下一个字符开始遍历。如果map中所有的值都为0，则找到了一个符合条件的子串，索引压入数组。

【考查内容】字符串

class Solution {
public:
void initializeMap(map<string,int>& map, vector<string>& words){
    for(int i = 0 ;i< words.size();i++){//初始化map
        if(map.count(words[i])==0){
            map[words[i]] = 1;
        }
        else
            map[words[i]] += 1;
    }
}
vector<int> findSubstring(string s, vector<string>& words) {
    map<string, int> mapOfVec;
    int singleWordLen = words[0].length();//单个字符串长度
    int wordsLen = words.size();
    int slen = s.length();
    int i,j,count;
    bool countChanged = false;//判断是否改变过map中的值，如果没变则无需重新初始化
    vector<int> result;
    count = wordsLen; //一个计数器表示还需要找到的字串个数
    if(wordsLen == 0 || slen ==0) return result;
    initializeMap(mapOfVec,words);
    for(i = 0; i<= slen-wordsLen*singleWordLen;i++){
        string subStr = s.substr(i,singleWordLen);// 取出字串
        j = i;
        while(mapOfVec.count(subStr)!=0 && mapOfVec[subStr]!=0 && j+singleWordLen<=slen){//当该字串存在于map中且值大于0，并且j不越界的情况下
            mapOfVec[subStr] -=1; //值减1
            count--;                      
            countChanged = true;   //改变了map的值
            j=j+singleWordLen;
            subStr = s.substr(j,singleWordLen); //下一个字串
            if(mapOfVec.count(subStr)==0){ 
            break;
            }
        }
        if(count==0){
            result.push_back(i); //找齐所有字符串数组中的字串后把该索引压入；
        }
        if(countChanged){ //若未找到而且改变了map的值需要重新初始化map和count
            mapOfVec.clear();
            initializeMap(mapOfVec,words);
            count = wordsLen;
            countChanged = false;
        }
    }
    return result;
    
}
};