problem:
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?
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题意:一个数组中,只有一个元素出现一次,其他元素出现K次,(k为奇数3)
thinking:
(1)这道题适用于出现奇次的情形,解法是:对数组所有数的每一位出现1的次数进行统计,对K取余后,就为待求数在该位的位数(0或者1),
再将2进制转换为10进制即可
(2)这道题我假设int为32位,有些机器不是32位。
code:
class Solution { public: int singleNumber(vector<int>& nums) { string s; int a=0x0001; int count=0; for(int i=0;i<32;i++) { for(int j=0;j<nums.size();j++) { if((nums[j]&a)!=0) count++; } s.push_back('0'+count%3); a=a<<1; count=0; } return reverse_string_to_int(s); } protected: int reverse_string_to_int(string s) { int a=1; int ret=0; for(int i=0;i<s.size();i++) { ret+=(s.at(i)-'0')*a; a*=2; } return ret; } };
时间: 2024-12-29 12:03:21