LeetCode-350. Intersection of Two Arrays II (Java)

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

Example:
Given nums1 = [1, 2, 2, 1], nums2 = [2, 2], return [2, 2].

Note:

• Each element in the result should appear as many times as it shows in both arrays.

• The result can be in any order.

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代码

public class Solution {
    public int[] intersect(int[] nums1, int[] nums2) {
         HashMap map = new HashMap();
        ArrayList result = new ArrayList();
        for(int i = 0; i < nums1.length; i++)
        {
            if(map.containsKey(nums1[i])) map.put(nums1[i], map.get(nums1[i])+1);
            else map.put(nums1[i], 1);
        }
    
        for(int i = 0; i < nums2.length; i++)
        {
            if(map.containsKey(nums2[i]) && map.get(nums2[i]) > 0)
            {
                result.add(nums2[i]);
                map.put(nums2[i], map.get(nums2[i])-1);
            }
        }
    
       int[] r = new int[result.size()];
       for(int i = 0; i < result.size(); i++)
       {
           r[i] = result.get(i);
       }
    
       return r;

    }
}

还有一种方法，在求重合时比较常用

public class Solution {
    public int[] intersect(int[] nums1, int[] nums2) {
            Arrays.sort(nums1);
            Arrays.sort(nums2);
            int pnt1 = 0;
            int pnt2 = 0;
            ArrayList myList = new ArrayList();
            while((pnt1 < nums1.length) && (pnt2 < nums2.length)){
                if(nums1[pnt1] < nums2[pnt2]) pnt1++;
                else if(nums1[pnt1] > nums2[pnt2]) pnt2++;
                else {
                    myList.add(nums1[pnt1]);
                    pnt1++; pnt2++;
                }
            }
            int len = myList.size();
            int[] res = new int[len];
            for(int i = 0; i < len; i++){
                res[i] = myList.get(i);
            }
            return res;
    }
}

使用while循环，然后对两个数组的索引进行操作，但是前提是这两个数组要是有序的。