Intersection of Two Arrays II两个数组交集（重要！）

https://leetcode.com/problems/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.

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?

思路是用map记录数和nums1中每个数出现的次数的映射； nums2中出现一次，这个次数减1

class Solution {
public:
	vector<int> intersect(vector<int>& nums1, vector<int>& nums2) {
		vector<int> res;
		if (nums1.size() == 0 || nums2.size() == 0){
			return res;
		}
		unordered_map<int,int> m;
		for (int n : nums1){
			m[n]++;//注意利用map这种索引取值的特性
		}
		for (int n : nums2){
			if (m.count(n) && m[n] > 0){
				res.push_back(n);
				m[n]--;
			}
		}
		return res;
	}
};

现在回答

• What if the given array is already sorted? How would you optimize your algorithm?

思路是用两个指针分别指向两个数组开头，相等则加入结果，两个指针都增加；     不等的话小的指针增加；    直到有一个指针达到末尾；

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