Intersection of Two Arrays II

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

Example

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.

Follow up:

• What if the given array is already sorted? How would you optimize your algorithm?

• What if nums1's size is small compared to nums2's size? Which algorithm is better?

• What if elements of nums2 are stored on disk, and the memory is limited such that you cannot load all elements into the memory at once?

Note

这里不同的结果需要是重复的

Code

使用排序加双指针

public int[] intersect2(int[] nums1, int[] nums2) {
    List<Integer> res = new ArrayList<Integer>();
    Arrays.sort(nums1);
    Arrays.sort(nums2);
    for(int i = 0, j = 0; i< nums1.length && j<nums2.length;){
            if(nums1[i]<nums2[j]){
                i++;
            }
            else if(nums1[i] == nums2[j]){
                res.add(nums1[i]);
                i++;
                j++;
            }
            else{
                j++;
            }
    }
    int[] result = new int[res.size()];
    for(int i = 0; i<res.size();i++){
        result[i] = res.get(i);
    }
    return result;
}

用counting map

public int[] intersect(int[] nums1, int[] nums2) {
    List<Integer> res = new ArrayList<Integer>();
    HashMap<Integer, Integer> map = new HashMap<>();

    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])){
            if(map.get(nums2[i]) > 0){
                res.add(nums2[i]);
                map.put(nums2[i], map.get(nums2[i]) - 1);
            }
        }
    }

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