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?
解法1：Hashmap
解法2：双指针
Python:
class Solution(object):
def intersect(self, nums1, nums2):
"""
:type nums1: List[int]
:type nums2: List[int]
:rtype: List[int]
"""
if len(nums1) > len(nums2):
return self.intersect(nums2, nums1)
lookup = collections.defaultdict(int)
for i in nums1:
lookup[i] += 1
res = []
for i in nums2:
if lookup[i] > 0:
res += i,
lookup[i] -= 1
return res
# If the given array is already sorted, and the memory is limited, and (m << n or m >> n).
# Time: O(min(m, n) * log(max(m, n)))
# Space: O(1)
# Binary search solution.
class Solution(object):
def intersect(self, nums1, nums2):
"""
:type nums1: List[int]
:type nums2: List[int]
:rtype: List[int]
"""
if len(nums1) > len(nums2):
return self.intersect(nums2, nums1)
def binary_search(compare, nums, left, right, target):
while left < right:
mid = left + (right - left) / 2
if compare(nums[mid], target):
right = mid
else:
left = mid + 1
return left
nums1.sort(), nums2.sort() # Make sure it is sorted, doesn't count in time.
res = []
left = 0
for i in nums1:
left = binary_search(lambda x, y: x >= y, nums2, left, len(nums2), i)
if left != len(nums2) and nums2[left] == i:
res += i,
left += 1
return res
# If the given array is already sorted, and the memory is limited or m ~ n.
# Time: O(m + n)
# Soace: O(1)
# Two pointers solution.
class Solution(object):
def intersect(self, nums1, nums2):
"""
:type nums1: List[int]
:type nums2: List[int]
:rtype: List[int]
"""
nums1.sort(), nums2.sort() # Make sure it is sorted, doesn't count in time.
res = []
it1, it2 = 0, 0
while it1 < len(nums1) and it2 < len(nums2):
if nums1[it1] < nums2[it2]:
it1 += 1
elif nums1[it1] > nums2[it2]:
it2 += 1
else:
res += nums1[it1],
it1 += 1
it2 += 1
return res
# If the given array is not sorted, and the memory is limited.
# Time: O(max(m, n) * log(max(m, n)))
# Space: O(1)
# Two pointers solution.
class Solution(object):
def intersect(self, nums1, nums2):
"""
:type nums1: List[int]
:type nums2: List[int]
:rtype: List[int]
"""
nums1.sort(), nums2.sort() # O(max(m, n) * log(max(m, n)))
res = []
it1, it2 = 0, 0
while it1 < len(nums1) and it2 < len(nums2):
if nums1[it1] < nums2[it2]:
it1 += 1
elif nums1[it1] > nums2[it2]:
it2 += 1
else:
res += nums1[it1],
it1 += 1
it2 += 1
return res
C++:
// If the given array is not sorted and the memory is unlimited.
// Time: O(m + n)
// Space: O(min(m, n))
// Hash solution.
class Solution {
public:
vector<int> intersect(vector<int>& nums1, vector<int>& nums2) {
if (nums1.size() > nums2.size()) {
return intersect(nums2, nums1);
}
unordered_map<int, int> lookup;
for (const auto& i : nums1) {
++lookup[i];
}
vector<int> result;
for (const auto& i : nums2) {
if (lookup[i] > 0) {
result.emplace_back(i);
--lookup[i];
}
}
return result;
}
};
C++:
// If the given array is already sorted, and the memory is limited, and (m << n or m >> n).
// Time: O(min(m, n) * log(max(m, n)))
// Space: O(1)
// Binary search solution.
class Solution {
public:
vector<int> intersect(vector<int>& nums1, vector<int>& nums2) {
if (nums1.size() > nums2.size()) {
return intersect(nums2, nums1);
}
// Make sure it is sorted, doesn't count in time.
sort(nums1.begin(), nums1.end());
sort(nums2.begin(), nums2.end());
vector<int> result;
auto it = nums2.cbegin();
for (const auto& i : nums1) {
it = lower_bound(it, nums2.cend(), i);
if (it != nums2.end() && *it == i) {
result.emplace_back(*it++);
}
}
return result;
}
};
C++:
// If the given array is already sorted, and the memory is limited or m ~ n.
// Time: O(m + n)
// Soace: O(1)
// Two pointers solution.
class Solution {
public:
vector<int> intersect(vector<int>& nums1, vector<int>& nums2) {
vector<int> result;
// Make sure it is sorted, doesn't count in time.
sort(nums1.begin(), nums1.end());
sort(nums2.begin(), nums2.end());
auto it1 = nums1.cbegin(), it2 = nums2.cbegin();
while (it1 != nums1.cend() && it2 != nums2.cend()) {
if (*it1 < *it2) {
++it1;
} else if (*it1 > *it2) {
++it2;
} else {
result.emplace_back(*it1);
++it1, ++it2;
}
}
return result;
}
};
C++:
// If the given array is not sorted, and the memory is limited.
// Time: O(max(m, n) * log(max(m, n)))
// Space: O(1)
// Two pointers solution.
class Solution {
public:
vector<int> intersect(vector<int>& nums1, vector<int>& nums2) {
vector<int> result;
// O(max(m, n) * log(max(m, n)))
sort(nums1.begin(), nums1.end());
sort(nums2.begin(), nums2.end());
auto it1 = nums1.cbegin(), it2 = nums2.cbegin();
while (it1 != nums1.cend() && it2 != nums2.cend()) {
if (*it1 < *it2) {
++it1;
} else if (*it1 > *it2) {
++it2;
} else {
result.emplace_back(*it1);
++it1, ++it2;
}
}
return result;
}
};
类似题目：
[LeetCode] 349. Intersection of Two Arrays 两个数组相交
[LeetCode] 160. Intersection of Two Linked Lists 求两个链表的交集