leetcode 349 Intersection of Two Arrays || leetcode 350 Intersection of Two Arrays II

leetcode Intersection of Two Arrays

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

Example: Given nums1 = [1, 2, 2, 1], nums2 = [2, 2], return [2].

Note:

• Each element in the result must be unique.
• The result can be in any order.

题目地址：leetcode Intersection of Two Arrays

题意：给定两个数组，求他们的交集（不算重复）

思路：太TM水了，直接上哈希表即可。。。

Python

class Solution(object):
    def intersection(self, nums1, nums2):
        """
        :type nums1: List[int]
        :type nums2: List[int]
        :rtype: List[int]
        """
        ans = []
        nums1 = set(nums1)
        for num in nums2:
            if num in nums1:
                ans.append(num)
        return list(set(ans)) #remove duplicate

 

当然也可以one line

class Solution(object):
    def intersection(self, nums1, nums2):
        """
        :type nums1: List[int]
        :type nums2: List[int]
        :rtype: List[int]
        """
        return list(set(nums1) & set(nums2))

C++

class Solution {
public:
	vector<int> intersection(vector<int>& nums1, vector<int>& nums2) {
		unordered_set<int> vis;
		unordered_set<int> intersection;
		vector<int> ans;
		for (int num : nums1) vis.insert(num);
		for (int num : nums2) if (vis.find(num) != vis.end()) intersection.insert(num);
		for (int num : intersection) ans.push_back(num);
		return ans;
	}
};

Java

class Solution {
    public int[] intersection(int[] nums1, int[] nums2) {
        HashSet<Integer> vis1 = new HashSet<Integer>();
        HashSet<Integer> intersection = new HashSet<Integer>();
        for(int num:nums1) vis1.add(num);
        for(int num:nums2) if(vis1.contains(num)) intersection.add(num);
        int []ans = new int[intersection.size()];
        int i = 0;
        for(int num:intersection)
            ans[i++] = num;
        return ans;
    }
}

 

---

leetcode Intersection of Two Arrays II

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 num2'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?

题目地址：Intersection of Two Arrays II

题意：给定两个数组，求他们的交集（可以重复）

思路：用个字典。。。计数即可。

class Solution(object):
    def intersect(self, nums1, nums2):
        """
        :type nums1: List[int]
        :type nums2: List[int]
        :rtype: List[int]
        """
        ans = []
        nums1 = collections.Counter(nums1)
        for x in nums2:
            if x in nums1:
                ans.append(x)
                nums1[x] -= 1
                if nums1[x] == 0:
                    del nums1[x]
        return ans

 

本题是leetcode 349 Intersection of Two Arrays  和 leetcode 350 Intersection of Two Arrays II题解

更多题解可以查看：https://www.hrwhisper.me/leetcode-algorithm-solution/