QUESTION:
Given an integer array nums and an integer k, return the number of pairs (i, j) where i < j such that |nums[i] - nums[j]| == k.
| The value of | x | is defined as: |
x if x >= 0. -x if x < 0.
Example 1:
Input: nums = [1,2,2,1], k = 1
Output: 4
Explanation: The pairs with an absolute difference of 1 are:
- [1,2,2,1]
- [1,2,2,1]
- [1,2,2,1]
- [1,2,2,1]
Example 2:
Input: nums = [1,3], k = 3
Output: 0
Explanation: There are no pairs with an absolute difference of 3.
Example 3:
Input: nums = [3,2,1,5,4], k = 2
Output: 3
Explanation: The pairs with an absolute difference of 2 are:
- [3,2,1,5,4]
- [3,2,1,5,4]
- [3,2,1,5,4]
Constraints:
1 <= nums.length <= 200
1 <= nums[i] <= 100
1 <= k <= 99
EXPLANATION:
简单题, 双循环比对绝对值就行. 注意边界情况.
SOLUTION:
class Solution {
func countKDifference(_ nums: [Int], _ k: Int) -> Int {
var result:Int = 0;
if nums.count == 1 {
return result
}
for i in 0...nums.count-2 {
for j in i+1 ... nums.count-1 {
if abs(nums[i] - nums[j]) == k {
result += 1
}
}
}
return result
}
}