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?

解法一:hashmap

public class Solution {
    public int[] intersect(int[] nums1, int[] nums2) {
        HashMap<Integer, Integer> map = new HashMap<Integer, Integer>();
        ArrayList<Integer> result = new ArrayList<Integer>();
        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;
    }
}

解法二:

Sort both arrays, use two pointers

Time complexity: O(nlogn)

public class Solution {
    public int[] intersect(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;
    }
}
时间: 2024-12-18 06:25:20

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