原文链接: https://leetcode-cn.com/problems/string-matching-in-an-array

英文原文

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.

中文题目

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。

通过代码

高赞题解

解题思路

两次循环
第一次遍历数组拼接成一个长字符串S
第二次遍历数组查找字符串在S中出现的位置，如果是别人的子串，那么在S中的出现次数一定 >= 2，那么起始跟结束的位置索引一定是不一样的，如果一样说明不是子串。
注意：为了避免前一字符串的尾部与后一字符串的头部构成混淆项，各字符串用,隔开拼接。

代码

class Solution {
    public List<String> stringMatching(String[] words) {
        List<String> list = new ArrayList<>();
        if (words.length == 0) return list;
        StringBuilder builder = new StringBuilder();
        for (int i = 0; i < words.length; i++){
            String str = words[i];
            builder.append(str + ",");
        }

        for (int i = 0; i < words.length; i++){
            String str = words[i];
            if (builder.toString().indexOf(str) != builder.toString().lastIndexOf(str)) list.add(str);
        }
        return list;
    }
}

统计信息

通过次数	提交次数	AC比率
13934	22668	61.5%

提交历史

提交时间	提交结果	执行时间	内存消耗	语言