Gives you an integer array nums. If a set of numbers (i, j) satisfies nums[i] == nums[j] and i <j, it can be considered as a good number pair. Returns the number of good pairs.
Example 1:
Input: nums = [1,2,3,1,1,3]
Output: 4
Explanation: There are 4 groups of good number pairs, namely (0,3), (0,4), (3,4), (2,5), and the subscript starts from 0
Example 2:
Input: nums = [1,1,1,1]
Output: 6
Explanation: Each group of numbers in the array is a good number pair
Example 3:
Input: nums = [1,2,3]
Output: 0
Algorithm 1: Use double-layer loop to directly traverse statistics, similar to bubble sort
void Fun(int *arr,int length)//o(n^2) o(1)
{
int num = 0;
for (int i = 0; i < length; ++i)
{
for (int j = i + 1; j < length; ++j)
{
if (arr[i] == arr[j])
{
++num;
}
}
}
printf("%d", num);
}
Algorithm 2: Use an iterator to traverse the array, equal to +1
int GoodNumberPair(vector<int>& nums)
{
int cnt = 0;
for (auto i = nums.begin(); i != nums.end(); i++) // 使用迭代器遍历
{
for (auto j = i+1; j!=nums.end(); j++) // 遍历
{
if (*i == *j) cnt++; // 相等加一
}
}
return cnt;
}
// 打印数组
void Print(vector<int>& vec)
{
for (auto v : vec)
{
cout << v << " ";
}
cout << endl;
}
Test code:
int main()
{
vector<int> test1{
1,2,3,1,1,3 };
auto ret1 = GoodNumberPair(test1);
vector<int> test2{
1,1,1,1 };
auto ret2 = GoodNumberPair(test2);
vector<int> test3{
1,2,3 };
auto ret3 = GoodNumberPair(test3);
Print(test1);
cout << ret1 << endl;
Print(test2);
cout << ret2 << endl;
Print(test3);
cout << ret3 << endl;
/*
int arr[] = {1,1,1,1};
int length = sizeof(arr) / sizeof(arr[0]);
Fun(arr, length);
*/
return 0;
}