1. 题目
[intersection-of-two-arrays](https://leetcode.com/problems/intersection-of-two-arrays/)
2. 算法
* set
* 两指针
* 二分
* stl
3. 代码
* set
class Solution {
public:
vector<int> intersection(vector<int>& nums1, vector<int>& nums2) {
set<int> s(nums1.begin(), nums1.end()), res;
for (auto a : nums2) {
if (s.count(a)) res.insert(a);
}
return vector<int>(res.begin(), res.end());
}
};
* 两指针
class Solution {
public:
vector<int> intersection(vector<int>& nums1, vector<int>& nums2) {
vector<int> res;
int i = 0, j = 0;
sort(nums1.begin(), nums1.end());
sort(nums2.begin(), nums2.end());
while (i < nums1.size() && j < nums2.size()) {
if (nums1[i] < nums2[j]) ++i;
else if (nums1[i] > nums2[j]) ++j;
else {
if (res.empty() || res.back() != nums1[i]) {
res.push_back(nums1[i]);
}
++i; ++j;
}
}
return res;
}
};
* 二分
class Solution {
public:
vector<int> intersection(vector<int>& nums1, vector<int>& nums2) {
set<int> res;
sort(nums2.begin(), nums2.end());
for (auto a : nums1) {
if (binarySearch(nums2, a)) {
res.insert(a);
}
}
return vector<int>(res.begin(), res.end());
}
bool binarySearch(vector<int> &nums, int target) {
int left = 0, right = nums.size();
while (left < right) {
int mid = left + (right - left) / 2;
if (nums[mid] == target) return true;
else if (nums[mid] < target) left = mid + 1;
else right = mid;
}
return false;
}
};
* stl
class Solution {
public:
vector<int> intersection(vector<int>& nums1, vector<int>& nums2) {
set<int> s1(nums1.begin(), nums1.end()), s2(nums2.begin(), nums2.end()), res;
set_intersection(s1.begin(), s1.end(), s2.begin(), s2.end(), inserter(res, res.begin()));
return vector<int>(res.begin(), res.end());
}
};