Given two arrays, write a function to compute their intersection.
Example 1:
Input: nums1 = [1,2,2,1], nums2 = [2,2]
Output: [2,2]
Example 2:
Input: nums1 = [4,9,5], nums2 = [9,4,9,8,4]
Output: [4,9]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?
LeetCode：链接
其实就是优化349题。不过这题是尽可能多的返回重复的数字。
第一种方法：用字典统计第一个列表都出现了哪些数及出现的次数，然后顺序遍历第二个列表，发现同时出现的数则加入到结果列表中，同时将第一个列表中相应的出现次数减一。
class Solution(object):
def intersect(self, nums1, nums2):
"""
:type nums1: List[int]
:type nums2: List[int]
:rtype: List[int]
"""
res = []
map = {}
for i in nums1:
map[i] = map[i]+1 if i in map else 1
for j in nums2:
if j in map and map[j] > 0:
res.append(j)
map[j] -= 1
return res第二种方法：排序加双指针。
class Solution(object):
def sort(self, nums, start, end):
if end <= start:
return start
index1, index2 = start, end
base = nums[start]
while start < end:
while start < end and nums[end] >= base:
end -= 1
nums[start] = nums[end]
while start < end and nums[start] <= base:
start += 1
nums[end] = nums[start]
nums[start] = base
self.sort(nums, index1, start-1)
self.sort(nums, start+1, index2)
def intersect(self, nums1, nums2):
"""
:type nums1: List[int]
:type nums2: List[int]
:rtype: List[int]
"""
self.sort(nums1, 0, len(nums1)-1)
self.sort(nums2, 0, len(nums2)-1)
i, j = 0, 0
res = []
while i < len(nums1) and j < len(nums2):
if nums1[i] < nums2[j]:
i += 1
elif nums1[i] > nums2[j]:
j += 1
else:
res.append(nums1[i])
i += 1
j += 1
return res Follow Up
Q1: What if the given array is already sorted? How would you optimize your algorithm?
A: 采用solution 2
Q2: What if nums1's size is small compared to nums2's size? Which algorithm is better?
A: 如果nums1相对于nums2非常小，那么把nums1做成哈希表，哈希表占用空间更小
Q3: 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?
A：如果只有nums2不能放在内存中，则将nums1做成哈希表，nums2分批加载到内存中处理。(If only nums2 cannot fit in memory, put all elements of nums1 into a HashMap, read chunks of array that fit into the memory, and record the intersections.)
如果nums1和nums2都很大，都不适合储存在内存，那么就用外部排序分别来sort它们。将每2G(举例)读入内存，使用2指针技术，然后从内存中读取更多的2G。重复此操作，直到没有更多数据从磁盘读取。(If both nums1 and nums2 are so huge that neither fit into the memory, sort them using external sort, read (let’s say) 2G of each into memory and then using the 2 pointer technique, then read 2G more from the array that has been exhausted. Repeat this until no more data to read from disk.)