追加字符以获得子序列

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.

/*
 * 思路：
 * 使用 Java Stream API 来解决这个问题。
 * 我们还是从 s 的开头和 t 的开头同时开始匹配，
 * 当匹配到 t 的结尾时，说明 t 已经是 s 的子序列。
 * 如果 s 遍历完都没有匹配到 t 的结尾，
 * 那么需要将 t 的剩余字符全部追加到 s 的末尾。
 */
import java.util.stream.IntStream;
public class Solution {
    public int appendCharacters(String s, String t) {
        int[] idx = {0};
        IntStream.range(0, s.length()).forEach(i -> {
            if (idx[0] < t.length() && s.charAt(i) == t.charAt(idx[0])) {
                idx[0]++;
            }
        });
        return t.length() - idx[0];
    }
}

题解的核心思想是通过一个指针来遍历字符串t，同时遍历字符串s。对于s中的每一个字符，如果它与t中当前指针所指向的字符相同，则移动t的指针到下一个字符。这一过程继续，直到s被完全遍历或者t的指针移动到字符串t的末尾。如果在遍历完s后，t的指针已经到达t的末尾，说明t已经是s的子序列，此时无需追加任何字符。否则，需要追加的字符数就是t的剩余部分的长度，即len(t) - cur_t。

O(n)

O(1)