Lintcode 1187. K-diff Pairs in an Array (Easy) (Python)

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