Substring with Concatenation of All Words

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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).

解题思路：将L中字符串极其个数建立字典，建立一个已经搜索的字符串与个数的字典并按位递归搜索S，当满足已经搜索个数与L个数一致时满足条件.若没有搜索到，或者搜索到的字符串个数超过建立的目标字典中的个数则跳过;

#include<vector>
#include<string>
#include<map>
#include<algorithm>

using namespace std;

bool search(string &S, int idx, int count, const int &length, const int &Num, map<string, int>&StringMap, map<string, int>&BeSearchMap)
{
	if (count == Num)
		return true;
	string StrSingle = S.substr(idx, length);
	if (StringMap.find(StrSingle) != StringMap.end())
	{
		if (++BeSearchMap[StrSingle] > StringMap[StrSingle])
			return false;
		return search(S, idx + length, count + 1, length, Num, StringMap, BeSearchMap);
	}
	else
		return false;
}
vector<int> findSubstring(string S, vector<string> &L) {
	vector<int>ResultIdxVector;
	map<string, int>StringMap;
	map<string, int>BeSearchMap;
	for_each(L.begin(), L.end(), [&StringMap](const string str){StringMap[str]++;});
	if (StringMap.empty())
		return ResultIdxVector;
	int length = L[0].size();
	int Num    = L.size();
	for (int i = 0; i != S.size()-length*Num+1;++i)
	{
		BeSearchMap.clear();//每次搜索都需要重置已搜索的字符串字典
		if (search(S, i, 0, length, Num, StringMap, BeSearchMap))
			ResultIdxVector.push_back(i);
	}
	return ResultIdxVector;
}

Substring with Concatenation of All Words

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原文地址：http://blog.csdn.net/li_chihang/article/details/43269965