【Leetcode】350. Intersection of Two Arrays II（求解数组的交集）

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 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?

题目大意：

给出两个数组求解数组的交集。

解题思路：

对两个数组进行由小到大的排序，我们标记两个数组的首位，分别判断当前的状态。

1.如果当前位置相等，idx1++，idx2++，ans.push_back；

2.如果当前位置idx1数值较大，则idx2++；

3.如果当前位置idx2数值较大，则idx1++；

class Solution {
public:
    vector intersect(vector& nums1, vector& nums2) {
        vector ans;
        sort(nums1.begin(), nums1.end());
        sort(nums2.begin(), nums2.end());
        int l,r;
        l = 0, r = 0;
        while(l nums2[r]){
                r++;
            }else{
                l++;
            }
        }
        return ans;
    }
};