532. K-diff Pairs in an Array

Medium
Given an array of integers nums and an integer k, return the number of unique k-diff pairs in the array.
A k-diff pair is an integer pair (nums[i], nums[j]), where the following are true:
  • 0 <= i, j < nums.length
  • i != j
  • nums[i] - nums[j] == k
Notice that |val| denotes the absolute value of val.
Example 1:
Input: nums = [3,1,4,1,5], k = 2
Output:
2
Explanation:
There are two 2-diff pairs in the array, (1, 3) and (3, 5).
Although we have two 1s in the input, we should only return the number of unique pairs.
Example 2:
Input: nums = [1,2,3,4,5], k = 1
Output:
4
Explanation:
There are four 1-diff pairs in the array, (1, 2), (2, 3), (3, 4) and (4, 5).
Example 3:
Input: nums = [1,3,1,5,4], k = 0
Output:
1
Explanation:
There is one 0-diff pair in the array, (1, 1).
Constraints:
  • 1 <= nums.length <= 10^4
  • -10^7 <= nums[i] <= 10^7
  • 0 <= k <= 10^7

解題

使用 map 記錄每個數字出現的次數,將 k == 0 的情況另外做處理。
func findPairs(nums []int, k int) int {
res := 0
m := make(map[int]int)
​
for _, n := range nums {
m[n]++
}
​
if k == 0 {
for key := range m {
if m[key] >= 2 {
res++
}
}
​
return res
}
​
for key := range m {
if m[key + k] >= 1 {
res++
}
}
​
return res
}
​
​