350. Intersection of Two Arrays II 难度：easy

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.

思路：

对数组nums1进行排序；对数组nums2进行排序；遍历数组nums1和nums2中元素，并比较对应的元素，若相等，则将其保存到输出结果中，并变化两个数组对应的索引，不等，则变化较小元素对应的索引即可。

程序：

class Solution {
public:
vector<int> intersect(vector<int>& nums1, vector<int>& nums2) {

sort(nums1.begin(), nums1.end());
sort(nums2.begin(), nums2.end());
vector<int> result;
for (int i = 0, j = 0; i < nums1.size() && j < nums2.size(); )
{
if (nums1[i] == nums2[j])
{
result.push_back(nums1[i]);
i++;
j++;
}
else if (nums1[i] < nums2[j])
i++;
else if (nums1[i] > nums2[j])
j++;
}
return result;
}
};