leetcode 30. 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 words exactly 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).

一眼看过去，以为是ac自动机。

后来看题目发现真难懂，看了看网上的解释，懂了。

题目意思转换一下就是求，words里面的单词排列组合成 k 求k 在s中出现的位置。

比如

"wordgoodgoodgoodbestword"
["word","good","best","good"]

words里面有4个单词，排列组合出24种字符串k。然后找出k在字符串中s出现的位置。

当然写代码时我们不能把24种组合都求出来，可以用map处理嘛。

时间复杂度O(n*m*len(s)) n*m为words中所有单词加起来的长度

class Solution {
public:
vector<int> findSubstring(string s, vector<string>& words) {
vector<int> ans;
int n = words.size();
if (n == 0) return ans;
int m = words[0].size();
unordered_map<string, int> mp;
for (int i = 0; i < n; ++i) {
mp[words[i]]++;
}
for (int i = 0; i + n*m <= s.size(); ++i) {
int num = 0;
unordered_map<string, int> mp2 = mp;
for (int j = 0; j < n; ++j) {
string tmp = s.substr(i + j*m, m);
if (mp2[tmp] > 0) num++;
mp2[tmp]--;
}
if (num != n) continue;
ans.push_back(i);
}
return ans;
}
};