Leetcode 350. Intersection of Two Arrays II 数组交集2 解题报告

1 解题思想

这道题和前两天的那个是一个系列的，只是那一个输出不能重复，而这一题重复了多少个，就输出多少个，所以先看下这个：
Leetcode 349. Intersection of Two Arrays 解题报告 Python Java

做法，在Java上，就是把原来的HashSet转变成HashMap用来计数就可以了，剩下就很简单了

2 原题

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?

3 AC解

public class Solution {
    public int[] intersect(int[] nums1, int[] nums2) {
         HashMap map = new HashMap();
        for(int i=0;iif(map.containsKey(nums1[i])==false)
                map.put(nums1[i],1);
            else
                map.put(nums1[i],1+map.get(nums1[i])); 
        List resultList = new ArrayList();
        for (int i=0;iif(map.containsKey(nums2[i])){
                if(map.get(nums2[i])>0){
                     resultList.add(nums2[i]);
                     map.put(nums2[i],-1+map.get(nums2[i])); 
                }
            }
        int result[] = new int[resultList.size()];
        for(int i=0;iget(i);
        return result;
    }
}