LeetCode 392. Is Subsequence

最新推荐文章于 2025-09-10 15:24:29 发布

转载最新推荐文章于 2025-09-10 15:24:29 发布 · 47 阅读

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原文链接：http://www.cnblogs.com/angryorange/p/5899923.html

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#数据结构与算法

本文介绍了一个简单的方法来检查一个字符串是否是另一个字符串的子序列。通过遍历较短字符串的每个字符，并在较长字符串中寻找这些字符出现的位置，以此验证子序列关系。文章提供了具体的Java代码实现。

Given a string s and a string t, check if s is subsequence of t.

You may assume that there is only lower case English letters in both s and t. t is potentially a very long (length ~= 500,000) string, and s is a short string (<=100).

A subsequence of a string is a new string which is formed from the original string by deleting some (can be none) of the characters without disturbing the relative positions of the remaining characters. (ie, "ace" is a subsequence of "abcde" while "aec" is not).

Example 1:
s = "abc", t = "ahbgdc"

Return true.

Example 2:
s = "axc", t = "ahbgdc"

Return false.

 1 // check if s is subsequence of t.
 2     public static boolean isSubsequence(String s, String t) {
 3         if (t.length() < s.length())
 4             return false;
 5         //　查找的启示位置
 6         int prev = 0;
 7         // 遍历s的字符
 8         for (int i = 0; i < s.length(); i++) {
 9             // 依次获取每个字符
10             char tempChar = s.charAt(i);
11             // 在上一次查找出字符的位置之后，查找t是否包含指定字符
12             prev = t.indexOf(tempChar, prev);
13             // t中不包含s的字符
14             if (prev == -1)
15                 return false;
16             // 下一次查找位置为：本次查找出字符的下一个位置
17             prev++;
18         }
19         return true;
20     }

题目考察如何判断s是t的字串
用到的API：
1. String.charAt(int i)：Returns the <code>char</code> value at the specified index.
2. String.indexOf(char , index)：Returns the index within this string of the first occurrence of the specified character, starting the search at the specified index.

转载于:https://www.cnblogs.com/angryorange/p/5899923.html