Intersection-of-Two-Arrays

发表于 2017-12-15 | 更新于： 2018-08-22

LeetCode#349 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.

解题思路

没什么说的，直接遍历查找，找到就存值，直接看代码。

解题代码【.CPP】

class Solution {
public:
    std::vector<int> intersection(std::vector<int>& nums1, std::vector<int>& nums2) {
        std::set<int> rset;
        for (auto it = nums1.begin() ; it != nums1.end() ; ++it){
            if(std::find(nums2.begin() , nums2.end() , *it) != std::end(nums2)){
                rset.insert(*it);
            }
        }
        return std::vector<int>(rset.begin(),rset.end());
    }
};

Intersection-of-Two-Arrays-II

发表于 2017-12-15 | 更新于： 2018-08-22

LeetCode#350 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 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?

解题思路

之前有一道题也是求两个数组的交点Intersection of Two Arrays，不过那个题比较简单，只需要循环然后使用std::find查找就可以了。这个题需要有重复的，我们可以将第一个数组的值和出现的次数存到map里，然后遍历第二个数组，如果在map里出现过就存入结果集里，并且将出现次数减一。

解题代码【.CPP】

class Solution {
public:
    vector<int> intersect(vector<int>& nums1, vector<int>& nums2) {
        unordered_map<int , int> um;
        for (auto num : nums1){
            if(um.find(num) != um.end()) ++um[num];
            else um[num] = 1;
        }
        vector<int> ret(0);
        for (auto num: nums2) {
            if(um[num] > 0){
                ret.push_back(num);
                --um[num];
            }
        }
        return ret;
    }
};