LeetCode 350. Intersection of Two Arrays II（数组交集）

原题网址：https://leetcode.com/problems/intersection-of-two-arrays-ii/

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?

方法一：哈希映射计数。

public class Solution {    public int[] intersect(int[] nums1, int[] nums2) {        Map<Integer, Integer> map1 = new HashMap<>();        for(int num: nums1) {            Integer count = map1.get(num);            if (count == null) count = 1; else count ++;            map1.put(num, count);        }        List<Integer> list2 = new ArrayList<>();        for(int num: nums2) {            Integer count = map1.get(num);            if (count == null) continue;            list2.add(num);            count --;            if (count == 0) map1.remove(num); else map1.put(num, count);        }        int[] result = new int[list2.size()];        for(int i=0; i<list2.size(); i++) result[i] = list2.get(i);        return result;    }}

方法二：先排序再查找。

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