原題是：
Given two arrays, write a function to compute their intersection.
***: 349.是該題的簡單版本沉唠，區(qū)別是結(jié)果不需要返回重復(fù)的結(jié)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：
這個(gè)題有兩個(gè)標(biāo)準(zhǔn)解法：
1.是我代碼所寫的two pointer方法。先將兩個(gè)數(shù)組排序，然后兩指針分別比較對應(yīng)位置。誰大了被饿，挪動(dòng)另一個(gè)指針。相等則把該元素插入結(jié)果中。
2.用hash table 存第一個(gè)數(shù)組的元素對應(yīng)元素出現(xiàn)的個(gè)數(shù)柔昼。再遍歷第二個(gè)數(shù)組，對于hashtable中已有的Key,填入結(jié)果炎辨，并將value值減去1捕透。
對follow-up：
3.如果內(nèi)存不夠，那只能用哈希的方法，第二個(gè)數(shù)組分批載入乙嘀。而第一個(gè)方法末购，由于需要排序，無法分批載入數(shù)組虎谢。
代碼是：
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
