264. Ugly Number II

• 思路1

出错代码：

 class Solution 
{
public:
    int nthUglyNumber(int n) 
	{
		if(n == 1)
		{
			return 1;
		}
		
        int i = 1;
		int flag = 1;
        while(1)
		{
			i++;
			if(isUgly(i))
			{
				flag++;
			}
			if(flag == n)
			{
				break;
			}
		}
		return i;
    }
	
private:
    bool isUgly(int n)
	{
		vector<int> nums;
		
		if(n == 3 || n == 5 || n == 2)
		{
			return true;
		}
		
		while(n)
		{
			nums.push_back(n % 10);
			n = n / 10;
		}
		
		if(nums[0] == 5 || nums[0] == 0)
		{
			//可以被5整除
			return isUgly(n / 5);
		}
		else if(is3Divisible(nums))
		{
			return isUgly(n / 3);
		}
		else
		{
			//能被2整除
			return isUgly(n / 2);
		}
		
		return false;
	}
	
	bool is3Divisible(vector<int>& nums)
	{
		int sum = 0;
		for(int i = 0;i < nums.size();i++)
		{
			sum = sum + nums[i];
		}
		
		if((sum % 3) == 0)
		{
			return true;
		}
		else
		{
			return false;
		}
	}
};

这次出错出在了

	bool isUgly(int n)
	{
		vector<int> nums;
		int number = n;
		if (n == 3 || n == 5 || n == 2)
		{
			return true;
		}

		while (n)
		{
			nums.push_back(n % 10);
			n = n / 10;
		}

		if (nums[0] == 5 || nums[0] == 0)
		{
			//可以被5整除
			return isUgly(number / 5);
		}
		else if (is3Divisible(nums))
		{
			return isUgly(number / 3);
		}
		else
		{
			//能被2整除
			return isUgly(number / 2);
		}

		return false;
	}

code2 代码出错 出错在

		else
		{
			//能被2整除
			return isUgly(number / 2);
		}

当number = 7时可以测出这个bug。
code3 修改后的代码，果然不出我所料，超时了。跑了几个数据都通过了，但是当n = 218时，leetcode提示Time Limit Exceeded。

class Solution
{
public:
	int nthUglyNumber(int n)
	{
		if (n == 1)
		{
			return 1;
		}

		int i = 1;
		int flag = 1;
		while (1)
		{
			i++;
			if (isUgly(i))
			{
				flag++;
			}
			if (flag == n)
			{
				break;
			}
		}
		return i;
	}

private:
	bool isUgly(int n)
	{
		vector<int> nums;
		int number = n;
		if (n == 3 || n == 5 || n == 2)
		{
			return true;
		}

		while (n)
		{
			nums.push_back(n % 10);
			n = n / 10;
		}

		if (nums[0] == 5 || nums[0] == 0)
		{
			//可以被5整除
			return isUgly(number / 5);
		}
		else if (is3Divisible(nums))
		{
			return isUgly(number / 3);
		}
		else if(is2Divisible(nums))
		{
			//能被2整除
			return isUgly(number / 2);
		}

		return false;
	}

	bool is3Divisible(vector<int>& nums)
	{
		int sum = 0;
		for (int i = 0; i < nums.size(); i++)
		{
			sum = sum + nums[i];
		}

		if ((sum % 3) == 0)
		{
			return true;
		}
		else
		{
			return false;
		}
	}

	bool is2Divisible(vector<int>& nums)
	{
		if (nums[0] % 2 == 0)
		{
			return true;
		}
		else
		{
			return false;
		}
	}
};

再一次出错的代码

class Solution
{
public:
	int nthUglyNumber(int n)
	{
		vector<int> ugly;
		ugly.push_back(1);

		int Multiply2 = 0;
		int Multiply3 = 0;
		int Multiply5 = 0;
		int i2 = 0;
		int i3 = 0;
		int i5 = 0;

		while (true)
		{
			if (n < ugly.size())
			{
				return ugly[n - 1];
			}
			else
			{
				Multiply2 = ugly[i2] * 2;
				Multiply3 = ugly[i3] * 3;
				Multiply5 = ugly[i5] * 5;

				int min = GetMin(Multiply2, Multiply3, Multiply5);
				ugly.push_back(min);

				if (min == Multiply2)
				{
					i2++;
				}
				else if (min == Multiply3)
				{
					i3++;
				}
				else
				{
					i5++;
				}

			}
		}

	}
private:
	int GetMin(int number1, int number2,int number3)
	{
		int min = (number1 < number2) ? number1 : number2;
		min = (min < number3) ? min : number3;
		return min;
	}
};

n = 10可以测出这个bug
最终AC版的code

class Solution
{
public:
	int nthUglyNumber(int n)
	{
		vector<int> ugly;
		ugly.push_back(1);

		int Multiply2 = 0;
		int Multiply3 = 0;
		int Multiply5 = 0;
		int i2 = 0;
		int i3 = 0;
		int i5 = 0;

		while (true)
		{
			if (n < ugly.size())
			{
				return ugly[n - 1];
			}
			else
			{
				Multiply2 = ugly[i2] * 2;
				Multiply3 = ugly[i3] * 3;
				Multiply5 = ugly[i5] * 5;

				int min = GetMin(Multiply2, Multiply3, Multiply5);
				ugly.push_back(min);

				if (min == Multiply2)
				{
					i2++;
				}
				if (min == Multiply3)
				{
					i3++;
				}
				if(min == Multiply5)
				{
					i5++;
				}

			}
		}

	}
private:
	int GetMin(int number1, int number2,int number3)
	{
		int min = (number1 < number2) ? number1 : number2;
		min = (min < number3) ? min : number3;
		return min;
	}
};