K-diff Pairs in an Array
Description:
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
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).
Code:
会报超时错误
class Solution:
"""
@param nums: an array of integers
@param k: an integer
@return: the number of unique k-diff pairs
"""
def findPairs(self, nums, k):
# Write your code here
l = len(nums)
if l <= 1:
return 0
nums.sort()
ind = 0
cnt = 0
if k == 0:
for i in range(l-1):
if nums[i+1]==nums[i]:
cnt+=1
return cnt
nums = list(set(nums))
l = len(nums)
for i in range(l-1):
ind = i+1
while ((ind < l) and (nums[ind]-nums[i])<=k):
if (nums[ind]-nums[i])==k:
cnt += 1
ind += 1
return cnt