Given two integer arrays nums1 and nums2, return an array of their intersection. Each element in the result must appear as many times as it shows in both arrays and you may return the result in any order.
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] Explanation: [9,4] is also accepted.
Constraints:
1 <= nums1.length, nums2.length <= 10000 <= nums1[i], nums2[i] <= 1000
Follow up:
nums1's size is small compared to nums2's size? Which algorithm is better?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);
List<Integer> intersection = new ArrayList<>();
for (int p1 = 0, p2 = 0; p1 < nums1.length && p2 < nums2.length;) {
if (nums1[p1] == nums2[p2]) {
intersection.add(nums1[p1]);
p1++;
p2++;
} else if (nums1[p1] < nums2[p2]) {
p1++;
} else
p2++;
}
return intersection.stream().mapToInt(Integer::intValue).toArray();
}
}class Solution {
public int[] intersect(int[] nums1, int[] nums2) {
Arrays.sort(nums1);
Arrays.sort(nums2);
List<Integer> intersection = new ArrayList<>();
for (int p1 = 0, p2 = 0; p1 < nums1.length && p2 < nums2.length;) {
if (nums1[p1] == nums2[p2]) {
intersection.add(nums1[p1]);
p1++;
p2++;
} else if (nums1[p1] < nums2[p2]) {
p1++;
} else
p2++;
}
return intersection.stream().mapToInt(Integer::intValue).toArray();
}
}