350. Intersection of Two Arrays II 数组交集2

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 num2'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?

给定两个数组，返回它们的共有部分，其中重复的元素也重复返回。

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【思路1】
将两个数组排序，然后顺序遍历并比较。
时间复杂度 O(max(m, n) log(max(m, n))) ，空间复杂度O(m + n)

class Solution {
public:
    vector intersect(vector& nums1, vector& nums2) {
        sort(nums1.begin(), nums1.end());
        sort(nums2.begin(), nums2.end());
        int n1 = (int)nums1.size(), n2 = (int)nums2.size();
        int i1 = 0, i2 = 0;
        vector res;
        while(i1 < n1 && i2 < n2){
            if(nums1[i1] == nums2[i2]) {
                res.push_back(nums1[i1]);
                i1++;
                i2++;
            }
            else if(nums1[i1] > nums2[i2]){
                i2++;
            }
            else{
                i1++;
            }
        }
        return res;
    }
};

【思路2】
利用关联容器map统计数字的数量。
时间复杂度 O(m + n)，空间复杂度O(m + n)。

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