【一天一道LeetCode】#350. Intersection of Two Arrays II

一天一道LeetCode

本系列文章已全部上传至我的github，地址：ZeeCoder‘s Github

欢迎大家关注我的新浪微博，我的新浪微博

欢迎转载，转载请注明出处

（一）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?

（二）解题

题目大意：找两个数组的交集，数组中有重复的数。

题目思路：【一天一道LeetCode】#349. Intersection of Two Arrays与上题差不多，只不过这道题不需要考虑去重的问题，所以，去掉重复检测那段就行。

class Solution {
public:
    vector<int> intersect(vector<int>& nums1, vector<int>& nums2) {
        vector<int> ret;
        if(nums1.empty()||nums2.empty()) return ret;
        sort(nums1.begin(),nums1.end());//首先对两个数组排序
        sort(nums2.begin(),nums2.end());
        auto iter1 = nums1.begin();
        auto iter2 = nums2.begin();
        auto end1 = nums1.end();
        auto end2 = nums2.end();
        while(iter1 != end1&&iter2!=end2)//其中一个遍历完就退出
        {
            if(*iter1<*iter2){//*iter2大就把iter1往后找
                ++iter1;
            }
            else if(*iter1>*iter2){//*iter1大就把iter2往后找
                ++iter2;
            }
            else {//找到了交集
                ret.push_back(*iter1);//存储交集
                ++iter1;
                ++iter2;
            }
        }
        return ret;
    }
};