Single Number I II III

Single Number

题目：

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?

解法一：Runtime：60ms。比较容易想到，采用map存储各个数极其次数，然后遍历出次数为1的那个即可。这种方法比较适用多种情况。

class Solution
{
public:
int singleNumber(vector<int>& nums)
{
map<int, int> m;
for (int i = 0; i != nums.size(); ++i)
{
m[nums[i]]++;
}
for (auto iter = m.begin(); iter != m.end(); ++iter)
if (iter->second == 1)
return iter->first;
}
};

解法二：Runtime：16ms。基本上是最优解法了，采用了异或运算。了解三点即可理解该算法：第一，所有数与0异或的结果都是本身，即0^a = a； 第二，异或满足交换率，即不在乎先后顺序，a^b^c^b^d^a^c=d。第三，两个相同的数异或结果为0，相异异或结果为1。据此，我们可以将数组中所有的数进行异或运算，这样相同的数会两两相消，最终剩下的就是我们所要的结果。

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

测试案列：

int main()
{
Solution s;
vector<int> vec = { 1,2,3,4,3,2,1 };
cout << s.singleNumber(vec) << endl;

return 0;
}

Single Number II

题目：

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?

解法一：Single Number I 中的第一种解法同样适用于本题，Runtime为：28ms

解法二：Runtime：20ms。在计算机中数字是以二进制的形式存储的，并且此处我们默认是Win32应用程序。那么每位只可能是0或1两种情况，如果一个数字连续出现3次的话那么对应位必然出现三次。根据题意，一个整形数组，只有一个元素出现一次，其余都出现三次。那么只需要把这个数组里的所有数都看称二进制的形式，然后把各个数第i th位的数相加的和对3取模所得的结果就是那个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)
{
int count[32] = { 0 };
int result = 0;
{
for (int j = 0; j != nums.size(); ++j)
{
count[i] += ((nums[j] >> i) & 1);
}
result |= ((count[i] % 3) << i);
}
return result;
}
};

解法三：这种方法应该是最优的了，但是水平有限这种方法没看懂。先暂时贴一下别人的代码吧。

class Solution
{
public:
int singleNumber(vector<int>& A)
{
int ones = 0, twos = 0, threes = 0;
for (int i = 0; i != A.size(); ++i)
{
twos |= ones & A[i];
ones ^= A[i];
threes = ones & twos;
ones &= ~threes;
twos &= ~threes;
}
return ones;
}
};

Single Number III

题目：

Given an array of numbers nums, in which exactly two elements appear only once and all the other elements appear exactly twice. Find the two elements that appear only once.

For example:

Given nums = [1, 2, 1, 3, 2, 5], return [3, 5].

Note:

The order of the result is not important. So in the above example, [5, 3] is also correct.

Your algorithm should run in linear runtime complexity. Could you implement it using only

constant space complexity?

解法一：Runtime：36ms。

class Solution
{
public:
vector<int> singleNumber(vector<int>& nums)
{
map<int, int> m;
for (int i = 0; i != nums.size(); ++i)
{
m[nums[i]]++;
}
vector<int> result;
for (auto i = m.begin(); i != m.end(); ++i)
if (i->second == 1)
result.push_back(i->first);
return result;
}
};

测试案列：

int main()
{
Solution s;
vector<int> vec = { 1, 2, 1, 3, 2, 5 };
vector<int> temp = s.singleNumber(vec);
for (auto i : temp)
cout << i << endl;

return 0;
}