LeetCode 350. Intersection of Two Array

350. Intersection of Two Array

一、问题描述

Given two arrays, write a function to compute their intersection.

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、输入输出

Example:
Given nums1 = [1, 2, 2, 1], nums2 = [2, 2], return [2, 2].

三、解题思路

• 和之前找交叉的那道题不同之处在于，这里需要按照具体出现了几次都输出出来。个人想法很直，把两个数组分别存储到一个map中，记录下每个元素出现的次数。假设一共有n个元素，那么时间复杂度是o(n) map的实现是红黑树，插入的复杂度是o(logn)。然后遍历这两个map找到相同的元素，取出现的个数较小的那个作为最终结果存储到vector中返回。

class Solution {public:    vector<int> intersect(vector<int>& nums1, vector<int>& nums2) {        map<int, int> dict1, dict2;        int n1 = nums1.size(), n2 = nums2.size();        for (int i = 0; i < n1; ++i) {            if(dict1.find(nums1[i]) == dict1.end()) dict1.insert(make_pair(nums1[i], 1));            else dict1[nums1[i]]++;        }        for (int j = 0; j < n2; ++j) {            if(dict2.find(nums2[j]) == dict2.end()) dict2.insert(make_pair(nums2[j], 1));            else dict2[nums2[j]]++;        }        vector<int> ret;        for (auto ite : dict1){            int key = ite.first;            int cnt = 0;            map<int, int>::iterator ite2;            if((ite2 = dict2.find(key)) != dict2.end()){                cnt = min(ite.second, ite2->second);            }            for (int i = 0; i < cnt; ++i) {                ret.push_back(key);            }        }        return ret;    }};