题目如下:
In a array
A
of size2N
, there areN+1
unique elements, and exactly one of these elements is repeated N times.Return the element repeated
N
times.Example 1:
Input: [1,2,3,3] Output: 3Example 2:
Input: [2,1,2,5,3,2] Output: 2Example 3:
Input: [5,1,5,2,5,3,5,4] Output: 5Note:
4 <= A.length <= 10000
0 <= A[i] < 10000
A.length
is even
解题思路:送分题。因为题目没有要求不能用额外的内存,所以我的方法是用字典保存每个数字出现的次数,从而找到出现N的数字。
代码如下:
class Solution(object): def repeatedNTimes(self, A): """ :type A: List[int] :rtype: int """ dic = {} res = 0 for i in A: dic[i] = dic.setdefault(i,0) + 1 if dic[i] == len(A)/2: res = i break return res
原文地址:https://www.cnblogs.com/seyjs/p/10168114.html
时间: 2024-11-09 04:44:51