[leetcode] 350. Intersection of Two Arrays II 解题报告

题目链接: https://leetcode.com/problems/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?

思路: 还是用hash表, 简单粗暴, 如果是有序的可以不用hash表, 同时扫描两个数组, 找相同的. 如果nums2在磁盘中, 用hash表无影响

代码如下:

class Solution {
public:
    vector intersect(vector& nums1, vector& nums2) {
        if(nums1.size()==0 || nums2.size()==0) return vector();
        vector result;
        unordered_map hash;
        for(auto val: nums1) hash[val]++;
        for(auto val: nums2)
        {
            if(hash.count(val) && hash[val]>0) result.push_back(val);
            hash[val]--;
        }
        return result;
    }
};

python

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

from compiler.ast import flatten
class Solution(object):
    def intersect(self, nums1, nums2):
        """
        :type nums1: List[int]
        :type nums2: List[int]
        :rtype: List[int]
        """
        return flatten([[x]*min(nums1.count(x), nums2.count(x)) for x in set(nums1)])