LeetCode: Single Number I && II

I title:

Given an array of integers, every element appears twice except for one. Find that single one.

Note:
Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory?

思路：异或

class Solution {
public:
int singleNumber(vector<int>& nums) {
int single = nums[0];
for (int i = 1 ;i < nums.size(); i++){
single ^= nums[i];
}
return single;
}
};

II

title：

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?

思路：

这里我们需要重新思考，计算机是怎么存储数字的。考虑全部用二进制表示，如果我们把 第 ith 个位置上所有数字的和对3取余，那么只会有两个结果 0 或 1 (根据题意，3个0或3个1相加余数都为0). 因此取余的结果就是那个 “Single Number”.

一个直接的实现就是用大小为 32的数组来记录所有 位上的和。

class Solution {
public:
int singleNumber(vector<int>& nums) {
vector<int> v(32,0);
int result = 0;
for (int i = 0; i < 32; i++){
for (int j = 0 ;j < nums.size(); j++){
if ((nums[j] >> i) & 1)
v[i]++;
}
result |= ((v[i] % 3) << i);
}
return result;
}
};

这个算法是有改进的空间的，可以使用掩码变量：

ones

代表第ith 位只出现一次的掩码变量

twos

代表第ith 位只出现两次次的掩码变量

threes

代表第ith 位只出现三次的掩码变量

假设在数组的开头连续出现3次5，则变化如下：

ones = 101
twos = 0
threes = 0
--------------
ones = 0
twos = 101
threes = 0
--------------
ones = 0
twos = 0
threes = 101
--------------

当第 ith 位出现3次时，我们就

ones

和

twos

的第 ith 位设置为0. 最终的答案就是 ones。

class Solution {
public:
int singleNumber(vector<int>& nums) {
int one = 0, two = 0, three = 0;
for (int i = 0; i < nums.size(); i++){
two |= (one & nums[i]);
one ^= nums[i];
three = one & two;
one &= ~three;
two &= ~three;
}
return one;
}
};