题目：
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).
开始想的维持一个窗口，c++实现代码如下：
vector<int> findSubstring(string S, vector<string> &L)
{
vector<int> result;
map<string, int> lmap;
int llen = L.size();
if(llen == 0)return result;
int l = L[0].length();
int i,j,k,start,window = 0;
//vector to map
lmap.clear();
vector<string>::iterator iter;
for(iter = L.begin(); iter != L.end(); iter++)
{
lmap.insert(pair<string,int>(*iter, -1));
}
//find
for(i = 0; i < l; i++)
{
start = i;
for(auto it = lmap.begin(); it != lmap.end(); ++it)
(*it).second = -1;
window = 0;
for(j = i; j < S.length(); j+=l)
{
if(j+l >= S.length())break;
map<string, int>::iterator itr = lmap.find(S.substr(j, l));
if(itr == lmap.end())
{
//如果没有匹配上，则前面的都不匹配，window=0，前面匹配的字符串对应的值都改为-1
for(k = start; k < j; k+=l)
{
lmap.find(S.substr(k,l)) -> second = -1;
}
start = j+l;
window = 0;
}else
{
int cur = itr ->second;
if(cur == -1)
{
itr -> second = j;
window++;
if(window == llen)
{
result.push_back(start);
window--;
lmap.find(S.substr(j-l*(llen - 1),l)) -> second = -1;
start += l;
}
}else
{
for(k = start; k <= cur; k+=l)
{
lmap.find(S.substr(k,l)) -> second = -1;
window--;
}
start += l;
itr->second = j;
window++;
}
}
}
}
return result;
}
};
但是有一个问题遇到L里有相同字符串时匹配不了。现在还没解决。
后来是参考了http://blog.csdn.net/ojshilu/article/details/22212703这篇博客。
思路是：假设L中的单位长度为n，依次从S中取长度为n的子串，如果在L中，就记下来。需要借助hash或map，如果整个L都匹配完了，就算是一个concatenation；当匹配错误的时候，S右移一个位置。但是这样效率很低，用了742ms，因为有些查找是可以跳过而没有跳过。还是想办法把窗口思想的实现吧。
vector<int> findSubstring(string S, vector<string> &L) {
vector<int> result;
int n = L[0].size();
int len = n * L.size();
if(len > S.size())
return result;
map<string, int> m;
for(int i=0;i<L.size();i++)
m[L[i]]++; // 发现可以不用特意初始化直接开始自增
int idx = 0;
map<string, int> tmp;
while(idx <= S.size() - len)
{
bool flag = true;
tmp.clear();
for(int i=idx;i<=idx+n*(L.size()-1);i+=n)
{
string now = S.substr(i, n);
if(m.find(now) == m.end())
{
flag = false;
break;
}
tmp[now]++;
if(tmp[now] > m[now])
{
flag = false;
break;
}
}
if(flag == true)
result.push_back(idx);
idx++;
}
return result;
}
[leetcode]Substring with Concatenation of All Words
原文：http://www.cnblogs.com/zhutianpeng/p/4276098.html