LeetCode392 判断是否子字符串

最新推荐文章于 2023-02-12 17:51:53 发布

fruit513

最新推荐文章于 2023-02-12 17:51:53 发布

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本文链接：https://blog.youkuaiyun.com/fruit513/article/details/84936960

本文介绍了一种检查字符串s是否为另一字符串t的子序列的算法。通过遍历s中的字符，利用indexOf方法在t中查找相应字符的位置，确保s中字符在t中的相对位置不变。适用于s长度小于等于100，t长度可达500,000的情况。

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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.

Follow up:
If there are lots of incoming S, say S1, S2, … , Sk where k >= 1B, and you want to check one by one to see if T has its subsequence. In this scenario, how would you change your code?

Credits:
Special thanks to @pbrother for adding this problem and creating all test cases.
思路：先看有没有第一个字母，然后看第一个字母的索引后有没有第二个字母……

public boolean isSubsequence(String s, String t) {
        if(s == null || s.length() == 0) return true;
        if(t.indexOf(s.charAt(0)) == -1) return false;
        int pos = t.indexOf(s.charAt(0));
        for(int i = 1; i < s.length(); i++){      
            int idx = t.indexOf(s.charAt(i), pos+1);
            if(idx == -1) return false;
            pos = idx;
        }
        return true;
    }