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<>题目：1408. 数组中的字符串匹配
<>题目内容
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：[]
<>解题思路 ：
这里我们用两个循环去遍历，用stringbuilder去连接字符串
第一个循环将所有的字符加入到builder中
第二个循环去对比字符串，如果字符串是子字符串那么一定会出现两次，
所以判断首次出现的位置和第二次出现的位置不同，就代表他是子字符串
解题代码如下：
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; } }
小总结：
锻炼算法能力是很需要坚持的
从最初的什么都不会到现在简单题轻松解答挑战中等题，时间会替你记下全部的努力
