LeetCode 350. Intersection of Two Arrays II

题目：
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?

思路：
这题和LeetCode 349. Intersection of Two Arrays思路一模一样，只需要将那题中的set换成multiset即可。具体思路见注释。

代码：

class Solution {
public:vector<int> intersect(vector<int>& nums1, vector<int>& nums2) {vector<int> result;multiset<int> setnums1(nums1.begin(),nums1.end());//将两个vector的数值放到set里去年重合的multiset<int> setnums2(nums2.begin(),nums2.end());for(set<int>::iterator iter1=setnums1.begin(),iter2=setnums2.begin();iter1!=setnums1.end()&&iter2!=setnums2.end();){//分别从两个multiset从头往后循环，如果都没有到结尾（有一个到结尾就跳出循环）if(*iter1==*iter2){//如果两个值相等result.push_back(*iter1);//这个值就为重合的++iter1;//iter1和iter2都后移++iter2;}else if(*iter1<*iter2){//如果*iter1值小，iter1后移++iter1;}else if(*iter1>*iter2){//如果*iter2值小，iter2后移++iter2;}}return result;}
};

输出结果： 6ms

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