LeetCode | 1408. String Matching
2020-04-12 本文已影响0人
Wonz
LeetCode 1408. String Matching in an Array数组中的字符串匹配【Easy】【Python】【字符串】
Problem
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.
问题
给你一个字符串数组 words ,数组中的每个字符串都可以看作是一个单词。请你按 任意 顺序返回 words 中是其他单词的子字符串的所有单词。
如果你可以删除 words[j] 最左侧和/或最右侧的若干字符得到 word[i] ,那么字符串 words[i] 就是 words[j] 的一个子字符串。
示例 1:
输入:words = ["mass","as","hero","superhero"]
输出:["as","hero"]
解释:"as" 是 "mass" 的子字符串,"hero" 是 "superhero" 的子字符串。
["hero","as"] 也是有效的答案。
示例 2:
输入:words = ["leetcode","et","code"]
输出:["et","code"]
解释:"et" 和 "code" 都是 "leetcode" 的子字符串。
示例 3:
输入:words = ["blue","green","bu"]
输出:[]
提示:
1 <= words.length <= 100
1 <= words[i].length <= 30
- words[i] 仅包含小写英文字母。
- 题目数据 保证 每个 words[i] 都是独一无二的。
思路
字符串
暴力,最后要去重。
时间复杂度: O(n^2)
空间复杂度: O(n)
Python3代码
from typing import List
import copy
class Solution:
def stringMatching(self, words: List[str]) -> List[str]:
n = len(words)
res = []
for i in range(n):
temp = copy.deepcopy(words)
temp.remove(words[i])
for j in range(n - 1):
if words[i] in temp[j]:
res.append(words[i])
return list(set(res))