# [LeetCode] 30. Substring with Concatenation of All Words 串联所有单词的子串

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.

Example 1:

Input:
s = "barfoothefoobarman",
words = ["foo","bar"]
Output: [0,9]
Explanation: Substrings starting at index 0 and 9 are "barfoor" and "foobar" respectively.
The output order does not matter, returning [9,0] is fine too.


Example 2:

Input:
s = "wordgoodgoodgoodbestword",
words = ["word","good","best","word"]
Output: []


class Solution {
public:
vector<int> findSubstring(string s, vector<string>& words) {
if (s.empty() || words.empty()) return {};
vector<int> res;
int n = words.size(), len = words[0].size();
unordered_map<string, int> wordCnt;
for (auto &word : words) ++wordCnt[word];
for (int i = 0; i <= (int)s.size() - n * len; ++i) {
unordered_map<string, int> strCnt;
int j = 0;
for (j = 0; j < n; ++j) {
string t = s.substr(i + j * len, len);
if (!wordCnt.count(t)) break;
++strCnt[t];
if (strCnt[t] > wordCnt[t]) break;
}
if (j == n) res.push_back(i);
}
return res;
}
};

class Solution {
public:
vector<int> findSubstring(string s, vector<string>& words) {
if (s.empty() || words.empty()) return {};
vector<int> res;
int n = s.size(), cnt = words.size(), len = words[0].size();
unordered_map<string, int> m1;
for (string w : words) ++m1[w];
for (int i = 0; i < len; ++i) {
int left = i, count = 0;
unordered_map<string, int> m2;
for (int j = i; j <= n - len; j += len) {
string t = s.substr(j, len);
if (m1.count(t)) {
++m2[t];
if (m2[t] <= m1[t]) {
++count;
} else {
while (m2[t] > m1[t]) {
string t1 = s.substr(left, len);
--m2[t1];
if (m2[t1] < m1[t1]) --count;
left += len;
}
}
if (count == cnt) {
res.push_back(left);
--m2[s.substr(left, len)];
--count;
left += len;
}
} else {
m2.clear();
count = 0;
left = j + len;
}
}
}
return res;
}
};

Github 同步地址：

https://github.com/grandyang/leetcode/issues/30

Minimum Window Substring

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

https://leetcode.com/problems/substring-with-concatenation-of-all-words/discuss/13656/An-O(N)-solution-with-detailed-explanation

https://leetcode.com/problems/substring-with-concatenation-of-all-words/discuss/13658/Easy-Two-Map-Solution-(C%2B%2BJava)

https://leetcode.com/problems/substring-with-concatenation-of-all-words/discuss/13664/Simple-Java-Solution-with-Two-Pointers-and-Map

LeetCode All in One 题目讲解汇总(持续更新中...)

posted @ 2015-05-22 00:07  Grandyang  阅读(22527)  评论(11编辑  收藏  举报