350. Intersection of Two Arrays II & Add to List 349. Intersection of Two Arrays 笔记

350. Intersection 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.
返回两个数组的公共元素，要是相同的有多个就返回多个。用一个哈希表统计第一个数组的每个数的个数，遍历第二个数组，哈希表里头有就保存到结果，并把哈希表里的这个数的个数减1。
代码如下：

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

349. Intersection of Two Arrays
     和上边的区别是重复的元素只输出一次。
     就在遍历第二个数组的时候，输出这个之后把哈希表里这个
     数的值置为0就行了。
     代码如下：

class Solution {
public:
    vector intersection(vector& nums1, vector& nums2) {
        unordered_map m;
        vector res;
        for(int i = 0; i < nums1.size(); i++)
        {
            m[nums1[i]]++;
        }
        for(int i = 0; i < nums2.size(); i++)
        {
            m[nums2[i]]--;
            if(m[nums2[i]] >= 0)
            {
                res.push_back(nums2[i]);
                m[nums2[i]] = 0;
            }
        }
        return res;
        }
};