Given an array of string words
. Return all strings in words
which is substring of another word in any order.
String words[i]
is substring of words[j]
, if can be obtained removing some characters to left and/or right side of words[j]
.
Example 1:
Input: words = ["mass","as","hero","superhero"] Output: ["as","hero"] Explanation: "as" is substring of "mass" and "hero" is substring of "superhero". ["hero","as"] is also a valid answer.
Example 2:
Input: words = ["leetcode","et","code"] Output: ["et","code"] Explanation: "et", "code" are substring of "leetcode".
Example 3:
Input: words = ["blue","green","bu"] Output: []
Constraints:
1 <= words.length <= 100
1 <= words[i].length <= 30
words[i]
contains only lowercase English letters.- It’s guaranteed that
words[i]
will be unique.
Solution: Brute Force
Time complexity: O(n^2)
Space complexity: O(1)
C++
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// Author: Huahua class Solution { public: vector<string> stringMatching(vector<string>& words) { vector<string> ans; const int n = words.size(); for (int i = 0; i < n; ++i) for (int j = 0; j < n; ++j) { if (i == j) continue; if (words[j].find(words[i]) != string::npos) { ans.push_back(words[i]); break; } } return ans; } }; |
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