找到两个数组的前缀公共数组

You are given two 0-indexed integer permutations A and B of length n.

A prefix common array of A and B is an array C such that C[i] is equal to the count of numbers that are present at or before the index i in both A and B.

Return the prefix common array of A and B.

A sequence of n integers is called a permutation if it contains all integers from 1 to n exactly once.

 

Example 1:

Input: A = [1,3,2,4], B = [3,1,2,4]
Output: [0,2,3,4]
Explanation: At i = 0: no number is common, so C[0] = 0.
At i = 1: 1 and 3 are common in A and B, so C[1] = 2.
At i = 2: 1, 2, and 3 are common in A and B, so C[2] = 3.
At i = 3: 1, 2, 3, and 4 are common in A and B, so C[3] = 4.

Example 2:

Input: A = [2,3,1], B = [3,1,2]
Output: [0,1,3]
Explanation: At i = 0: no number is common, so C[0] = 0.
At i = 1: only 3 is common in A and B, so C[1] = 1.
At i = 2: 1, 2, and 3 are common in A and B, so C[2] = 3.

 

Constraints:

• 1 <= A.length == B.length == n <= 50
• 1 <= A[i], B[i] <= n
• It is guaranteed that A and B are both a permutation of n integers.

/*
题目思路：
1. 创建一个长度为 n 的结果数组 C。
2. 使用两个集合来跟踪 A 和 B 的前缀元素。
3. 遍历数组，比较当前索引之前的前缀元素，统计公共元素的数量并存储到 C 中。
*/

import java.util.*;
import java.util.stream.*;

public class Solution {
    public int[] findPrefixCommonArray(int[] A, int[] B) {
        int n = A.length;
        int[] C = new int[n];
        Set<Integer> setA = new HashSet<>();
        Set<Integer> setB = new HashSet<>();

        IntStream.range(0, n).forEach(i -> {
            setA.add(A[i]);
            setB.add(B[i]);
            C[i] = (int) setA.stream().filter(setB::contains).count();
        });
        return C;
    }
}

题解的核心思路在于利用两个集合来跟踪数组A和B中已经见过的元素。对于每个索引i，我们检查数组A和B的当前元素是否已经在对方的集合中出现过。如果是，相应的计数器（count_A或count_B）增加。这两个计数器分别跟踪A的元素在B的前缀中出现的次数和B的元素在A的前缀中出现的次数。最后，将这两个计数器的和存储在结果数组的相应位置。这种方法确保了即便元素以不同的顺序出现，也能正确计算出前缀中共有元素的数量。

O(n)

O(n)