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本文介绍两种计算两个数组交集的算法实现，包括确保每个元素只出现一次的交集及考虑重复次数的交集。提供了多种C++实现方案，如排序、双指针法和哈希表等。

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Intersection of Two Arrays
- 题目描述
- 解题
Intersection of Two Arrays II
- 题目描述
- 代码实现

349. Intersection of Two Arrays

题目描述

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

击败50%+的代码。

class Solution {
public:
    vector<int> intersection(vector<int>& nums1, vector<int>& nums2) {
        sort(nums1.begin(), nums1.end());
        sort(nums2.begin(), nums2.end());
        vector<int> rel;

        int len1 = nums1.size();
        int len2 = nums2.size();
        int stt2 = 0;

        for(int i = 0; i < len1; i++) {
            for(int j = stt2; j < len2; j++) {
                if(nums1[i] == nums2[j]) {
                    rel.push_back(nums1[i]);
                    while(j + 1 < len2 && nums2[j] == nums2[j+1])
                        j++;
                    stt2 = j + 1;
                }
            }

            while(i+1 < len1 && nums1[i] == nums1[i+1])
                i++;
        }

        return rel;
    }
};

在discussion发现了更加高级简洁的写法：

class Solution {
public:
    vector<int> intersection(vector<int>& nums1, vector<int>& nums2) {
        unordered_set<int> m(nums1.begin(), nums1.end());
        vector<int>rel;

        for(auto ele:nums2) {
            if(m.count(ele)) {
                rel.push_back(ele);
                m.erase(ele);
            }
        }

        return rel;
    }
};

350. Intersection of Two Arrays II

题目描述

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?

代码实现

class Solution {
public:
    vector<int> intersect(vector<int>& nums1, vector<int>& nums2) {
        sort(nums1.begin(), nums1.end());
        sort(nums2.begin(), nums2.end());
        vector<int> rel;

        int len1 = nums1.size();
        int len2 = nums2.size();
        int stt2 = 0;

        for(int i = 0; i < len1; i++) {
            for(int j = stt2; j < len2; j++) {
                if(nums1[i] == nums2[j]) {
                    rel.push_back(nums1[i]);
                    stt2 = j + 1;
                    break;
                }
            }
        }

        return rel;        
    }
};

全部调用标准库函数，击败了90%的c++代码：

class Solution {
public:
    vector<int> intersect(vector<int>& nums1, vector<int>& nums2) {
        sort(nums1.begin(), nums1.end());
        sort(nums2.begin(), nums2.end());
        a.erase(set_intersection(nums1.begin(), nums1.end(), nums2.begin(), nums2.end(), nums1.begin()), nums1.end());
        return nums1;       
    }
};

使用unordered_map，可以得到：

class Solution {
public:
    vector<int> intersect(vector<int>& nums1, vector<int>& nums2) {
        unordered_map<int, int> m;
        for(auto a:nums1) 
            m[a]++;

        vector<int> rel;    
        for(auto b:nums2)
            if((m[b]--) > 0)
                rel.push_back(b);

        return rel;       
    }
};