【LeetCode with Python】 Single Number II
2014-09-21 17:52
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原题页面:https://oj.leetcode.com/problems/single-number-ii/
题目类型:位运算
难度评价:★★★
本文地址:/article/1377467.html
Given an array of integers, every element appears three times except for one. Find that single one.
Note:
Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory?
与Single Number相比,前者是每个元素出现两次(除了一个特殊元素),而本题是每个元素出现三次(除了一个特殊元素)。
int数据共有32位,可以用32变量存储这N个元素中各个二进制位上1出现的次数,最后在进行模三操作,如果为1,那说明这一位是要找元素二进制表示中为1的那一位。但是这种做法估计不是最好的,应该有更好的算法,思考中。
原题页面:https://oj.leetcode.com/problems/single-number-ii/
题目类型:位运算
难度评价:★★★
本文地址:/article/1377467.html
Given an array of integers, every element appears three times except for one. Find that single one.
Note:
Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory?
与Single Number相比,前者是每个元素出现两次(除了一个特殊元素),而本题是每个元素出现三次(除了一个特殊元素)。
int数据共有32位,可以用32变量存储这N个元素中各个二进制位上1出现的次数,最后在进行模三操作,如果为1,那说明这一位是要找元素二进制表示中为1的那一位。但是这种做法估计不是最好的,应该有更好的算法,思考中。
class Solution: # @param A, a list of integer # @return an integer def singleNumber(self, A): len_A = len(A) counts = [0] * 32 res = 0 for i in range(0, 32): for num in A: #num = long(num) # it will make time limit exceeded counts[i] += (num>>i) & 1 res |= ((counts[i] % 3) << i) if 0 != counts[31] % 3: res -= pow(2, 32) ###### return res
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