392. Is Subsequence

最新推荐文章于 2024-03-08 17:49:31 发布

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最新推荐文章于 2024-03-08 17:49:31 发布

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本文介绍了一种检查字符串s是否为另一字符串t的子序列的高效算法，该算法的时间复杂度为O(m)+O(n)，其中m和n分别是字符串s和t的长度。文章通过示例展示了如何实现这一算法，并探讨了当面对大量输入字符串时，如何优化代码。

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?

时间复杂度O（m） + O(n)，程序如下所示：

class Solution {
    public boolean isSubsequence(String s, String t) {
        int lenS = s.length(), lenT = t.length();
        int index = 0, cnt = 0;
        for (int i = 0; i < lenS; ++ i){
            for (int j = index; j < lenT; ++ j){
                index ++;
                if (t.charAt(j) == s.charAt(i)){
                    cnt ++;
                    break;
                }
            }
        }
        return lenS == cnt;
    }
}