LEETCODE--Intersection of Two Arrays II

最新推荐文章于 2024-09-08 16:57:09 发布

prettysky123

最新推荐文章于 2024-09-08 16:57:09 发布

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本文介绍两种计算两个数组交集的方法。方法一采用map记录元素出现次数，遍历第二个数组并减少对应计数，实现复杂交集计算。方法二则先对两个数组排序，再用双指针技术找出相同元素。

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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?

方法一：

初次使用map这种key-value对应的容器。
参考c++ map的使用

class Solution {
public:
    vector<int> intersect(vector<int>& nums1, vector<int>& nums2) {
        int len1 = nums1.size();
        map<int, int> dict;
        for(int i = 0; i < len1; i++){
            ++dict[nums1[i]];
        }
        int len2 = nums2.size();
        vector<int> vec;
        for(int j = 0; j < len2; j++){
            if(dict[nums2[j]] != 0){
                vec.push_back(nums2[j]);
                dict[nums2[j]]--;
            }
        }
        return vec;
    }
};

方法二：
先排序后利用two point进行查找

class Solution {
public:
    vector<int> intersect(vector<int>& nums1, vector<int>& nums2) {
        int len1 = nums1.size();
        int len2 = nums2.size();
        vector<int> vec;
        sort(nums1.begin(),nums1.end());
        sort(nums2.begin(),nums2.end());
        int i = 0;
        int j = 0;
        while(i < len1 && j < len2){
            if(nums1[i] == nums2[j]){
                vec.push_back(nums1[i]);
                i++;
                j++;
            }else if(nums1[i] > nums2[j]){
                j++;
            }else{
                i++;
            }
        }
        return vec;
    }
};