[LeetCode] 532. K-diff Pairs in an Array

轉自LeetCode

Given an array of integers and an integer k, you need to find the number of unique k-diff pairs in the array. Here a k-diff pair is defined as an integer pair (i, j), where i and j are both numbers in the array and their absolute difference is k.

Example 1:

Input: [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:[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: [1, 3, 1, 5, 4], k = 0
Output: 1
Explanation: There is one 0-diff pair in the array, (1, 1).

Note:

1. The pairs (i, j) and (j, i) count as the same pair.
2. The length of the array won't exceed 10,000.
3. All the integers in the given input belong to the range: [-1e7, 1e7].

<Solution>

想法如下(參考解答)

• 對於每個數，都將它存到一個 set 裡面，用來檢查某數是否存在 array 中

• 同時，對於每個數，也去檢查它 +k 和 -k 的值，是否已經存在 set 裡面。如果存在，就代表 array 裡面有個數，可以配對成答案

• 答案用一個 set 儲存，存的值都是配對答案裡面數值較小的值，此作法是用來濾掉重複的答案

code 如下

C++

Java