[leetcode] 30. Substring with Concatenation of All Words 解题报告

题目链接：https://leetcode.com/problems/substring-with-concatenation-of-all-words/

You are given a string, s, and a list of words, words, that are all of the same length. Find all starting indices of substring(s) in s that is a concatenation
of each word in wordsexactly once and without any intervening characters.

For example, given:

s:

"barfoothefoobarman"

words:

["foo", "bar"]

You should return the indices:

[0,9]

.

(order does not matter).

思路：一种比较粗暴的方式就是先对words计数，然后在s的每个位置开始枚举固定的长度，如果能找到每个单词那就把当前的索引加到结果集合中．其实这样效率还是可以接受的．时间复杂度是O(n*k)，其中ｎ是s的长度，k是总共words的长度．

代码如下：

class Solution {
public:
vector<int> findSubstring(string s, vector<string>& words) {
if(s.size()==0 || words.size()==0) return {};
vector<int> result;
unordered_map<string, int> hash;
for(auto val: words) hash[val]++;
int m = words.size(), n = words[0].size(), len = s.size();
if(len < m*n) return {};
for(int i = 0, j; i <= len - m*n; i++)
{
unordered_map<string, int> find;
for(j = i; j < i + m*n; j+= n)
{
string tem = s.substr(j, n);
find[tem]++;
if(!hash.count(tem) || find[tem] > hash[tem]) break;
}
if(j >= i+m*n) result.push_back(i);
}
return result;
}
};