264. Ugly Number II

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• 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;
    }
};

264. Ugly Number II

标签：vat   limit   for   code   div   数据   number   flag   sum   

原文地址：https://www.cnblogs.com/Manual-Linux/p/12033432.html