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 num2’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?

解题思路

两个map分别存储两个数组的元素出现次数，然后以最少出现次数决定是否出现在结果数组中以及出现在结果数组中的次数。

实现代码

// Runtime: 13 ms
public class Solution {
    public int[] intersect(int[] nums1, int[] nums2) {
        Map<Integer, Integer> map1 = new HashMap<Integer, Integer>();
        for (int num : nums1) {
            if (map1.containsKey(num)) {
                map1.put(num, map1.get(num) + 1);
            } else {
                map1.put(num, 1);
            }
        }

        Map<Integer, Integer> map2 = new HashMap<Integer, Integer>();
        for (int num : nums2) {
            if (map2.containsKey(num)) {
                map2.put(num, map2.get(num) + 1);
            } else {
                map2.put(num, 1);
            }
        }

        List<Integer> list = new ArrayList<Integer>();
        for (Map.Entry<Integer, Integer> entry : map1.entrySet()) {
            int times = entry.getValue();
            if (map2.containsKey(entry.getKey())) {
                times = Math.min(times, map2.get(entry.getKey()));
            } else {
                times = 0;
            }
            for (int i = 0; i < times; i++) {
                list.add(entry.getKey());
            }
        }

        int[] res = new int[list.size()];
        for (int i = 0; i < list.size(); i++) {
            res[i] = list.get(i);
        }
        return res;
    }
}