[LeetCode] 2486. Append Characters to String to Make Subsequence

You are given two strings s and t consisting of only lowercase English letters.

Return the minimum number of characters that need to be appended to the end of s so that t becomes a subsequence of s.

A subsequence is a string that can be derived from another string by deleting some or no characters without changing the order of the remaining characters. 

Example 1:

Input: s = "coaching", t = "coding"
Output: 4
Explanation: Append the characters "ding" to the end of s so that s = "coachingding".
Now, t is a subsequence of s ("coachingding").
It can be shown that appending any 3 characters to the end of s will never make t a subsequence.

Example 2:

Input: s = "abcde", t = "a"
Output: 0
Explanation: t is already a subsequence of s ("abcde").

Example 3:

Input: s = "z", t = "abcde"
Output: 5
Explanation: Append the characters "abcde" to the end of s so that s = "zabcde".
Now, t is a subsequence of s ("zabcde").
It can be shown that appending any 4 characters to the end of s will never make t a subsequence.

Constraints:

• 1 <= s.length, t.length <= 105
• s and t consist only of lowercase English letters.

追加字符以获得子序列。

给你两个仅由小写英文字母组成的字符串 s 和 t 。

现在需要通过向 s 末尾追加字符的方式使 t 变成 s 的一个 子序列 ，返回需要追加的最少字符数。

子序列是一个可以由其他字符串删除部分（或不删除）字符但不改变剩下字符顺序得到的字符串。

来源：力扣（LeetCode）
链接：https://leetcode.cn/problems/append-characters-to-string-to-make-subsequence
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思路是双指针。这是判断子序列的基本题，我们用两个指针 i, j 分别指向 s 和 t 并同时开始遍历。当 s 遍历完的时候，如果 t 也遍历完了，说明 t 本身就是 s 的子序列，否则需要补足的字母个数 = n - j。

时间O(m) - s 的长度

空间O(1)

Java实现

 1 class Solution {
 2     public int appendCharacters(String s, String t) {
 3         int m = s.length();
 4         int n = t.length();
 5         int i = 0;
 6         int j = 0;
 7         while (i < m && j < n) {
 8             if (s.charAt(i) == t.charAt(j)) {
 9                 j++;
10             }
11             i++;
12         }
13         if (j == n) {
14             return 0;
15         }
16         return n - j;
17     }
18 }

LeetCode 题目总结