LeetCode350.Intersection of Two Arrays II(两个数组的交集 II)

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

Example 1:

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

Example 2:

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

Note:

  • Each element in the result should appear as many times as it shows in both arrays.
  • The result can be in any order.

给定两个数组,写一个方法来计算它们的交集。

例如: 
给定 nums1 = [1, 2, 2, 1], nums2 = [2, 2], 返回 [2, 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[] intersect(int[] nums1, int[] nums2) {
		TreeMap<Integer, Integer> map = new TreeMap<>();
		for (int num : nums1) {
			if (!map.containsKey(num))
				map.put(num, 1);
			else
				map.put(num, map.get(num) + 1);
		}
		ArrayList<Integer> res = new ArrayList<>();
		for (int num : nums2) {
			if (map.containsKey(num)) {
				res.add(num);
				map.put(num, map.get(num) - 1);
				if (map.get(num) == 0)
					map.remove(num);
			}
		}
		int[] ret = new int[res.size()];
		for (int i = 0; i < res.size(); i++)
			ret[i] = res.get(i);
		return ret;
	}
}

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.intersect(a, b)));
	}
}

 运行结果

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转载自blog.csdn.net/qq_26891141/article/details/85224965