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:
- The pairs (i, j) and (j, i) count as the same pair.
- The length of the array won't exceed 10,000.
- All the integers in the given input belong to the range: [-1e7, 1e7].
分三种情况讨论, 如果k< 0, return 0, 如果k == 0 , 看如果某个数至少有两个, +1, 如果k >0 , 利用Two sum的方法去找.
Code
class Solution(object): def findPairs(self, nums, k): """ :type nums: List[int] :type k: int :rtype: int """ if k < 0: return 0 if k==0: return sum(val > 1 for val in collections.Counter(nums).values()) nums, d, ans = set(nums), set(), 0 for num in nums: if num -k in d: ans += 1 if num + k in d: ans += 1 d.add(num) return ans