Article Directory
For all the LeetCode solution indexes, you can read this article - [Algorithm and Data Structure] LeetCode Solution .
1. Topic
Second, the solution
Idea analysis: Use a hash array to record the numbers that appear in nums1, and then traverse nums2 to find the value of 1 in the hash array, which is the element of the intersection.
The procedure is as follows：
class Solution {
public:
vector<int> intersection(vector<int>& nums1, vector<int>& nums2) {
unordered_set<int> result;
int hashTable[1005] = {
0 }; // 哈希数组
for(int num : nums1) hashTable[num] = 1; // nums1记录出现的数字
for (int num : nums2) {
if (hashTable[num]) {
result.insert(num);
}
}
return vector<int>(result.begin(), result.end());
}
};
Complexity analysis:
- time complexity: O ( m + n ) O(m+n) O(m+n ) , m and n are respectively divided into the lengths of the two arrays, and the program traverses the two arrays respectively.
- Space complexity: O ( t ) O(t) O ( t ) , wheret = min ( m , n ) t = min(m, n)t=min(m,n ) , the largest intersection of two arrays may be the shorter array.
3. Complete code
# include <iostream>
# include <unordered_set>
# include <vector>
# include <string>
using namespace std;
class Solution {
public:
vector<int> intersection(vector<int>& nums1, vector<int>& nums2) {
unordered_set<int> result;
int hashTable[1005] = {
0 }; // 哈希数组
for(int num : nums1) hashTable[num] = 1; // nums1记录出现的数字
for (int num : nums2) {
if (hashTable[num]) {
result.insert(num);
}
}
return vector<int>(result.begin(), result.end());
}
};
void VectorGenerator(int arr[], int len, vector<int>& nums1) {
for (int i = 0; i < len; i++) {
nums1.push_back(arr[i]);
}
}
void my_print(vector <int>& v, string msg)
{
cout << msg << endl;
for (vector<int>::iterator it = v.begin(); it != v.end(); it++)
{
cout << *it << " ";
}
cout << endl;
}
int main()
{
int arr1[] = {
4,9,5 };
int arr2[] = {
9,4,9,8,4 };
int arr1_len = sizeof(arr1) / sizeof(int);
int arr2_len = sizeof(arr2) / sizeof(int);
vector<int> nums1, nums2;
VectorGenerator(arr1, arr1_len, nums1);
VectorGenerator(arr2, arr2_len, nums2);
my_print(nums1, "nums1:");
my_print(nums2, "nums2:");
Solution s1;
vector<int> result = s1.intersection(nums1, nums2);
my_print(result, "result:");
system("pause");
return 0;
}
end