350. Intersection of Two Arrays II*-CFANZ编程社区

350. Intersection of Two Arrays II*

https://leetcode.com/problems/intersection-of-two-arrays-ii/

题目描述

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 ifnums1's size is small compared to nums2's size? Which algorithm is better?
What if elements ofnums2 are stored on disk, and the memory is limited such that you cannot load all elements into the memory at once?

C++ 实现 1

20210322 更新: 代码详解见 C++ 实现 2. 这次更新主要是哈希表记录元素较少的数组中的元素. 另外 C++ 实现 3 提供了利用排序来做的实现.

class Solution {
public:
    vector<int> intersect(vector<int>& nums1, vector<int>& nums2) {
        if (nums1.size() > nums2.size()) std::swap(nums1, nums2);
        unordered_map<int, int> record;
        for (auto &a : nums1) record[a] ++;
        vector<int> res;
        for (auto &a : nums2) {
            if (record.count(a)) {
                res.push_back(a);
                record[a] --;
                if (record[a] == 0) record.erase(a);
            }
        }
        return res;
    }
};

C++ 实现 2

用一个哈希表. 先统计 nums1 中的元素以及个数, 保存到 freq 中, 然后遍历 nums2, 判断每个元素是否在 freq 中, 同时还要保证对应的个数不为 0, 因为每次从 nums2 中找到了交集中的元素, freq 对应的个数需要减 1.

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

        vector<int> res;
        for (const auto &d : nums2) {
            if (freq.find(d) != freq.end() && freq[d] != 0) {
                res.push_back(d);
                freq[d] --;
            }
        }
        return res;
    }
};

C++ 实现 3

使用排序来实现. 对两个数组从小到大排好序后, 然后使用 i 和 j 分别来遍历两个数组. 如果两个元素相等, 那么就将该元素加入 res 中; 如果不相等, 那么只需要移动指向更小元素的索引即可.

class Solution {
public:
    vector<int> intersect(vector<int>& nums1, vector<int>& nums2) {
        std::sort(nums1.begin(), nums1.end());
        std::sort(nums2.begin(), nums2.end());
        int i = 0, j = 0;
        vector<int> res;
        while (i < nums1.size() && j < nums2.size()) {
            if (nums1[i] == nums2[j]) {
                res.push_back(nums1[i]);
                ++ i, ++ j;
            } else if (nums1[i] < nums2[j]) ++ i;
            else ++ j;
        }
        return res;
    }
};

C++ 实现 4

用两个哈希表的做法.

class Solution {
public:
    vector<int> intersect(vector<int>& nums1, vector<int>& nums2) {
        vector<int> res;
        unordered_map<int, int> set1, set2;
        for (auto &n : nums1) set1[n] ++;
        for (auto &n : nums2) set2[n] ++;
        for (auto &p : set1) {
            if (set2.count(p.first)) {
                int n = std::min(set2[p.first], p.second);
                auto tmp = vector<int>(n, p.first);
                std::copy(tmp.begin(), tmp.end(), back_inserter(res));
            }
        }
        return res;
    }
};