注意：凡是以英文出现的，都是题目提供的，包括答案代码里的前几行。
题目：
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?
解析：
这一题的思想也可以用来解决前面一题。首先对两个数组排序，然后分别用两个下标遍历数组，找出相同的部分添加到result数组即可。
如果要解决前面一题，在最后一个else分支添加一句while (i < nums1.length && nums1[i] == nums2[j]) i++;
答案：(Java)
public class Solution {
public int[] intersect(int[] nums1, int[] nums2) {
Arrays.sort(nums1);
Arrays.sort(nums2);
int[] result = new int[nums2.length];
int i = 0, j = 0, k = 0;
while ( i < nums1.length && j < nums2.length) {
if ( nums1[i] < nums2[j] ) {
i++;
}
else if ( nums1[i] > nums2[j] ) {
j++;
}
else {
result[k++] = nums1[i++];
// add to here
j++;
}
}
return Arrays.copyOfRange(result, 0, k);
}
}
