leetcode-350. Intersection of Two Arrays II

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.

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?

public class Solution {
    public int[] intersect(int[] nums1, int[] nums2) {
        Map<Integer,Integer> nums1_map = new HashMap<Integer,Integer>();
	List<Integer> ans = new ArrayList<Integer>();
	for(int num : nums1){
		
		if(nums1_map.containsKey(num)){
			nums1_map.put(num,nums1_map.get(num)+1);
		}else	nums1_map.put(num,1);
	}
	for(int num: nums2){
		if(nums1_map.isEmpty()) break;
		Integer value = nums1_map.get(num);
		if(value != null){
			ans.add(num);
			if(value == 1) 	
				nums1_map.remove(num);
			else 	
				nums1_map.put(num,value-1);

		}
	}
	int[] return_array = new int[ans.size()];
	for(int i = 0 ; i < ans.size(); i ++)
		return_array[i] = ans.get(i);
	return return_array;
    }
}

总结：使用map无论是否有序，时间效率其实差不多。不过第二点如果先判断谁长谁短，把短的先存到map里面可能会好很多。关于内存放不下，可以比如num1分成两部分，num2分成两部分，先把num1和num2放在内存中且交集的踢出去，然后轮换。