Description
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.
分析
题目的意思是：给定一个字符串数组，然后输出其中一个字符串是另一个字符串子串的集合。思路也很直接，把字符串数组按照长度进行排序，然后写个双循环判断字符串是否是另一个字符串的子串就行了。
代码
class Solution:
def stringMatching(self, words: List[str]) -> List[str]:
wd=sorted(words,key=len)
n=len(wd)
res=[]
for i in range(n):
for j in range(i+1,n):
if(wd[i] in wd[j]):
res.append(wd[i])
break
return res
参考文献
[LeetCode] [Python 3] String Matching in an Array. beats 98%
