leetcode刷题，总结，记录，备忘 350

leetcode350Intersection 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?

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与Intersection of Two Arrays的第一题比较相似，就是允许重复的，稍微改下代码就行了。

class Solution {
public:
    vector intersect(vector& nums1, vector& nums2) {
        sort(nums1.begin(), nums1.end());
        sort(nums2.begin(), nums2.end());
        
        vector result;
        
        int size1 = nums1.size();
        int size2 = nums2.size();
        
        int i = 0;
        int j = 0;
        
        while (i != size1 && j != size2)
        {
            if (nums1[i] < nums2[j])
            {
                i++;
            }
            else if (nums1[i] > nums2[j])
            {
                j++;
            }
            else if (nums1[i] == nums2[j])
            {
                result.push_back(nums1[i]);
                i++;
                j++;
            }
            else
            {
                i++;
                j++;
            }
        }
        
        return result;
    }
};

我还看到一个比较巧妙的解决方法，用map记录第一个数组中每个数字出现的次数，然后在第二个数组中通过数字做键，将数字出现次数做自减，根据结果是否小于0来判断是否出现在2个数组中。

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