349. Intersection of Two Arrays
Given two arrays, write a function to compute their intersection.
Example:
Given nums1 = [1, 2, 2, 1]
, nums2 = [2, 2]
, return [2]
.
Note:
-
Each element in the result must be unique.
-
The result can be in any order.
题目大意:
将两个数组中一样的元素存入结果数组返回。结果数组中的元素不能重复。
思路:
1.将数组1,数组2分别放入set中去重。
2.使用迭代器iterator遍历set1,在set2中找与set1相同的元素,找到就添加到结果数组中。
代码如下:
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class
Solution {
public
:
vector<
int
> intersection(vector<
int
>& nums1, vector<
int
>& nums2) {
vector<
int
> result;
set<
int
> set1;
set<
int
> set2;
set<
int
>::iterator it;
for
(
int
i = 0 ; i < nums1.size();i++)
if
(set1.find(nums1[i]) == set1.end())
set1.insert(nums1[i]);
for
(
int
i = 0 ; i < nums2.size();i++)
if
(set2.find(nums2[i]) == set2.end())
set2.insert(nums2[i]);
for
(it = set1.begin();it != set1.end();it++)
{
if
(set2.find(*it) != set2.end() )
result.push_back(*it);
}
return
result;
}
};
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本文转自313119992 51CTO博客,原文链接:http://blog.51cto.com/qiaopeng688/1837597