Leetcode350. Intersection of Two Arrays II

题目描述：

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?

题目分析：

这道题和http://blog.youkuaiyun.com/qq_27896185/article/details/53383882349题差不多，只有最后得出的数组允许出现重复的元素，并且需要有序。具体实现如下：

public class Solution {
        public int[] intersect(int[] nums1, int[] nums2) {
            List<Integer> list = new ArrayList<>();
            List<Integer> interList = new ArrayList<>();
            for (int i : nums1) {
                list.add(i);
            }
            for (int i = 0; i < nums2.length; i++) {
                if (list.contains(nums2[i])) {
                    interList.add(nums2[i]);
                    list.remove(list.indexOf(nums2[i]));
                }
            }
            int[] result = new int[interList.size()];
            int index = 0;
            for (int i : interList) {
                result[index++] = i;
            }
            return result;
        }   
    }

时间复杂度为O(n),空间复杂度为O(n);