350. Intersection of Two Arrays II

标签：==   NPU   turn   nts   com   memory   list   tor   cannot   

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

Example 1:

Input: nums1 = [1,2,2,1], nums2 = [2,2]
Output: [2,2]

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?

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

350. Intersection of Two Arrays II

标签：==   NPU   turn   nts   com   memory   list   tor   cannot   

原文地址：https://www.cnblogs.com/wentiliangkaihua/p/11770482.html