Given two arrays, write a function to compute their intersection.

Example 1:

Input: nums1 = [1,2,2,1], nums2 = [2,2]
Output: [2]

Example 2:

Input: nums1 = [4,9,5], nums2 = [9,4,9,8,4]
Output: [9,4]

Note:

• Each element in the result must be unique.
• The result can be in any order.

给定两个数组，写一个函数来计算它们的交集。

例子: 

给定 num1= [1, 2, 2, 1], nums2 = [2, 2], 返回 [2].

注意：

• 输出结果中每个元素出现的次数，应与元素在两个数组中出现的次数一致。
• 我们可以不考虑输出结果的顺序。

public class Solution {
	public int[] intersection(int[] nums1, int[] nums2) {

		TreeSet<Integer> set = new TreeSet<>();
		for (int num : nums1)
			set.add(num);

		ArrayList<Integer> list = new ArrayList<>();
		for (int num : nums2) {
			if (set.contains(num)) {
				list.add(num);
				set.remove(num);
			}
		}

		int[] res = new int[list.size()];
		for (int i = 0; i < list.size(); i++)
			res[i] = list.get(i);
		return res;

	}
}

测试代码

public class Test {
	public static void main(String[] args) {
		int[] a = { 1, 2, 2, 1 };
		int[] b = { 2, 2 };
		Solution solution = new Solution();
		System.out.println(Arrays.toString(solution.intersection(a, b)));
	}
}

运行结果