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 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 good number pairs, namely (0,3), (0,4), (3,4 ), (2,5), 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
Method 1: Brute force method
int numIdenticalPairs(int* nums, int numSize) // (时间复杂度O(n^2) 空间复杂度O(1))
{
int count = 0;//统计好数的数目
for (int i = 0; i < numSize; i++)
{
for (int j = i + 1; j < numSize; j++)
{
if (nums[i] == nums[j])
{
count++;
}
}
}
return count;
}
Method 2: Hash table method
int numIdenticalPairs(int* nums, int numsSize)
{
int count = 0;//累积好数对的数目
int hash[100] = {
0 };
for (int i = 0; i < numsSize; i++)
{
hash[nums[i]]++;//每个数目出现的次数
count += hash[nums[i]] - 1;//如果只出现一次好数对数目为0,出现的次数大于1则依次累加
}
return count;
}