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
}