LeetCode Substring with Concatenation of All Words

问题网址：https://leetcode.com/problems/substring-with-concatenation-of-all-words/description/

问题描述：

给你一个字符串s和一个长度相同的单词列表。 找出s中所有子字符串的起始索引，这些字符串中的每个单词只是一个字符串，没有任何中介字符。

例如，给出：

s：“barfoothefoobarman”

词语：[“foo”，“bar”]

你应该返回索引：[0,9]。

（顺序无关紧要）。

一个O(N)的解法

// travel all the words combinations to maintain a window
// there are wl(word len) times travel
// each time, n/wl words, mostly 2 times travel for each word
// one left side of the window, the other right side of the window
// so, time complexity O(wl * 2 * N/wl) = O(2N)
vector<int> findSubstring(string S, vector<string> &L) {
vector<int> ans;
int n = S.size(), cnt = L.size();
if (n <= 0 || cnt <= 0) return ans;

// init word occurence
unordered_map<string, int> dict;
for (int i = 0; i < cnt; ++i) dict[L[i]]++;

// travel all sub string combinations
int wl = L[0].size();
for (int i = 0; i < wl; ++i) {
int left = i, count = 0;
unordered_map<string, int> tdict;
for (int j = i; j <= n - wl; j += wl) {
string str = S.substr(j, wl);
// a valid word, accumulate results
if (dict.count(str)) {
tdict[str]++;
if (tdict[str] <= dict[str])
count++;
else {
// a more word, advance the window left side possiablly
while (tdict[str] > dict[str]) {
string str1 = S.substr(left, wl);
tdict[str1]--;
if (tdict[str1] < dict[str1]) count--;
left += wl;
}
}
// come to a result
if (count == cnt) {
ans.push_back(left);
// advance one word
tdict[S.substr(left, wl)]--;
count--;
left += wl;
}
}
// not a valid word, reset all vars
else {
tdict.clear();
count = 0;
left = j + wl;
}
}
}

return ans;
}

4000