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?

本题和intersection of two arrays1比较类似，只不过允许出现重复的元素了，那么能否用之前的三种方法来解决这个呢？二分查找那个肯定不可以，因为不可以动态删除数组元素；数组排序应该是可以的；另外时间复杂度O（n）的那个方法也是可以的，先说时间复杂度O（n）这个，只要把hashset换成hashmap就可以了，代码如下：

public class Solution {
    public int[] intersect(int[] nums1, int[] nums2) {
        Map<Integer,Integer> map = new HashMap<Integer,Integer>();
        List<Integer> res = new ArrayList<Integer>();
        for(int i:nums1){
            map.put(i,map.getOrDefault(i,0)+1);
        }
        for(int i:nums2){
            if(map.containsKey(i)&&map.get(i)>0){
                //find the int that appears in nums1,this method avoid remove operation
                map.put(i,map.get(i)-1);
                res.add(i);
            }
        }
        int[] result = new int[res.size()];
        int k=0;
        for(Integer i:res){
            result[k++] = i;
        }
        return result;
    }
}
//the run time complexity costs O(max(m,n)),the space complexity O(min(m,n));

 

数组排序代码如下：

public class Solution {
    public int[] intersect(int[] nums1, int[] nums2) {
        Map<Integer,Integer> map = new HashMap<Integer,Integer>();
        Arrays.sort(nums1);
        Arrays.sort(nums2);
        int count=  0;
        int i=0;
        int j=0;
        while(i!=nums1.length&&j!=nums2.length){
            if(nums1[i]<nums2[j]){
                i++;
            }else if(nums1[i]>nums2[j]){
                j++;
            }else{
                count++;
                map.put(nums1[i],map.getOrDefault(nums1[i],0)+1);
                i++;
                j++;
            }
        }
        int[] res = new int[count];
        int v = 0;
        for(int key:map.keySet()){
            for(int k=0;k<map.get(key);k++){
                res[v++] = key;
            }
        }
        return res;
    }
}

 本题的follow up的第三个问题有点难度，要看链接：

https://discuss.leetcode.com/topic/45992/solution-to-3rd-follow-up-question