[easy+][Array][Two-pointers]350. Intersection of Two Arrays II

原题是：

Given two arrays, write a function to compute their intersection.

***: 349.是该题的简单版本，区别是结果不需要返回重复的结果。
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?

思路：

这个题有两个标准解法：
1.是我代码所写的two pointer方法。先将两个数组排序，然后两指针分别比较对应位置。谁大了，挪动另一个指针。相等则把该元素插入结果中。
2.用hash table 存第一个数组的元素对应元素出现的个数。再遍历第二个数组，对于hashtable中已有的Key,填入结果，并将value值减去1。

对follow-up：
3.如果内存不够，那只能用哈希的方法，第二个数组分批载入。而第一个方法，由于需要排序，无法分批载入数组。

代码是：

class Solution:
    def intersect(self, nums1, nums2):
        """
        :type nums1: List[int]
        :type nums2: List[int]
        :rtype: List[int]
        """
        i,j = 0,0
        num1 = sorted(nums1)
        num2 = sorted(nums2)
        res = []
        
        while i < len(num1) and j < len(num2):
            if num1[i] < num2[j]:
                i += 1
            elif num1[i] > num2[j]:
                j += 1
            else:
                res.append(num1[i])
                i += 1
                j += 1
        return res