LeetCode 350. Intersection of Two Arrays II （两个数组的相交之二）

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?

 

 

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题目标签：Hash Table

　　题目给了我们两个array， 让我们找到相交的数字，这一题与 #349 一样，只是可以包括重复的数字。

　　所以利用HashMap 把 nums1 的数字和 出现次数存入；

　　比较nums2 和 map1 把相交的数字 和 次数 存入 intersect；

　　最后把intersect 里的 数字 按照它的出现次数 存入 int[] res。

 

 

 

Java Solution:

Runtime beats 23.67% 

完成日期：06/05/2017

关键词：HashMap

关键点：利用两个HashMap

class Solution 
{
    public int[] intersect(int[] nums1, int[] nums2) 
    {
        HashMap<Integer, Integer> map1 = new HashMap<>();
        HashMap<Integer, Integer> intersect = new HashMap<>();
        int[] res;
        int len = 0;
        int pos = 0;
        
        // store nums1 numbers into map1
        for(int n: nums1)
            map1.put(n, map1.getOrDefault(n, 0) + 1);

        
        // compare nums2 with map1, store intersected numbers into intersect
        for(int n: nums2)
        {
            if(map1.containsKey(n))
            {
                intersect.put(n, intersect.getOrDefault(n, 0) + 1);
                len++;
                
                map1.put(n, map1.get(n) - 1);
                
                if(map1.get(n) == 0)
                    map1.remove(n);
            }
                
        }
        
        res = new int[len];
        
        for(int n: intersect.keySet())
        {
            for(int i=0; i<intersect.get(n); i++)
                res[pos++] = n;
        }
        
        return res;
    }
}

参考资料：N/A

 

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