题目如下:
Given an array of integers
arr
of even lengthn
and an integerk
.We want to divide the array into exactly
n / 2
pairs such that the sum of each pair is divisible byk
.Return True If you can find a way to do that or False otherwise.
Example 1:
Input: arr = [1,2,3,4,5,10,6,7,8,9], k = 5 Output: true Explanation: Pairs are (1,9),(2,8),(3,7),(4,6) and (5,10).Example 2:
Input: arr = [1,2,3,4,5,6], k = 7 Output: true Explanation: Pairs are (1,6),(2,5) and(3,4).Example 3:
Input: arr = [1,2,3,4,5,6], k = 10 Output: false Explanation: You can try all possible pairs to see that there is no way to divide arr into 3 pairs each with sum divisible by 10.Example 4:
Input: arr = [-10,10], k = 2 Output: trueExample 5:
Input: arr = [-1,1,-2,2,-3,3,-4,4], k = 3 Output: trueConstraints:
arr.length == n
1 <= n <= 10^5
n
is even.-10^9 <= arr[i] <= 10^9
1 <= k <= 10^5
解题思路:统计arr中每个元素除k后的余数出现的次数,接下来只要判断余数i出现的次数是否和余数k-i出现的次数相等即可。对于余数为0的情况,只要满足出现的次数为偶数即可。
代码如下:
class Solution(object): def canArrange(self, arr, k): """ :type arr: List[int] :type k: int :rtype: bool """ dic = {} for i in arr: remainder = i % k dic[remainder] = dic.setdefault(remainder,0) + 1 for key,val in dic.iteritems(): if val == 0:continue elif key == 0 and val % 2 != 0: return False elif key == 0 and val % 2 == 0: continue elif k - key not in dic or dic[k-key] != val: return False return True