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).

解决该问题的关键是理解清楚要求。

给定一个目标字符串s，一个单词集合words。

要求使得words集合中所有元素连续出现在s中的首位置组成的集合（元素顺序不考虑）。

正如所给实例，目标字符串s: “barfoothefoobarman”

对比单词集合words: [“foo”, “bar”]

我们发现，在pos=0 ~ 5时“barfoo”恰好匹配，则0压入结果vector；

在pos=9 ~ 14时“foobar”恰好匹配，则9压入结果vector；

vector<int> findSubstring(string s, vector<string>& words) {
if (words.empty())return vector<int>();

vector<int> res;
unordered_map<string, int> word_cnt;

for (auto x : words)word_cnt[x]++;

int word_size = words[0].size();
int word_nums = words.size();
int s_size = s.size();

int i = 0, j = 0;
int len = s_size - word_size * word_nums + 1;
for (i = 0; i < len; i++){
unordered_map<string, int> temp_count;
for (j = 0; j < word_nums; j++){
string word = s.substr(i + j * word_size, word_size);
if (word_cnt.find(word) != word_cnt.end()){
++temp_count[word];
if (temp_count[word] > word_cnt[word])break;
}
else break;
}
if (j == word_nums)res.push_back(i);
}
return res;
}