标签：ddd app tag element bigger alt ons put array
https://leetcode.com/problems/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 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:
class Solution(object): def findPairs(self, nums, k): """ :type nums: List[int] :type k: int :rtype: int """ nums.sort() count = [] dict = {} for i in range(len(nums)): if nums[i] in dict: count.append((dict[nums[i]],nums[i])) dict[nums[i]+k] = nums[i] return len(set(count))
count = [(22,33),(66,66),(66,66)]
print len(count) ==> 3
print len(set(count)) ==> 2
Sol 2:
Use count method to track the appearance of elements.
class Solution(object): def findPairs(self, nums, k): """ :type nums: List[int] :type k: int :rtype: int """ res = 0 c = collections.Counter(nums) for i in c: if (k > 0 and i + k in c) or (k == 0 and c[i] > 1): res += 1 return res
Note:
1 collestions.counter is a python built-in method to count the appearance of elements.
import collections nums = ‘aaaaaaaaabbcccccddddeeef‘ c = collections.Counter(nums) print c
Counter({‘a‘: 9, ‘c‘: 5, ‘d‘: 4, ‘e‘: 3, ‘b‘: 2, ‘f‘: 1})
for i in c: print i
a c b e d f
标签：ddd app tag element bigger alt ons put array
原文地址：http://www.cnblogs.com/prmlab/p/7010984.html
