从1到n整数中1出现的次数：O(logn)算法

最新推荐文章于 2022-02-22 18:14:58 发布

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最新推荐文章于 2022-02-22 18:14:58 发布

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分类专栏：剑指offer 文章标签：算法 class

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问题：
给定一个十进制正整数N，写下从1开始，到N的所有整数，然后数一下其中出现的所有“1”的个数。
例如：
N= 2，写下1，2。这样只出现了1个“1”。
N= 12，我们会写下1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12。这样，1的个数是5。
一、暴力求解，不被推荐

class Solution
{
public:
    int NumberOf1Between1AndN( unsigned int n )
    {
        int number = 0;
        for ( int i = 0; i <= n; i++ )
        {
            number += NumberOf1( i );
        }
        return number;
    }
    int NumberOf1( unsigned int n )
    {
        int number = 0;
        while ( n )
        {
            if ( n % 10 == 1 )
                number++;
            n = n / 10;
        }
        return number;
    }
};

二、从数字规律着手解决
但是这种思路比较难以理解，推荐第三种。

class Solution
{
public:
    int NumberOf1Between1AndN( int n )
    {
        if ( n <= 0 )
            return 0;
        char strN[50];
        sprintf( strN, "%d", n );
        return NumberOf1( strN );
    }

    int NumberOf1( const char *strN )
    {
        if ( !strN || *strN < '0' || *strN > '9' || *strN == '\0' )
            return 0;
        int first = *strN - '0';
        unsigned int length = static_cast<unsigned int>( strlen( strN ) );

        if ( length == 1 && first == 0 )
            return 0;
        if ( length == 1 && first > 0 )
            return 1;

        int numFirstDigit = 0;
        if ( first > 1 )
            numFirstDigit = PowerBase10( length-1 );
        else if ( first == 1 )
            numFirstDigit = atoi( strN+1 ) + 1;

        int numOtherDigits = first * ( length-1 ) * PowerBase10( length-2 ); //有些疑惑
        int numRecursive = NumberOf1( strN+1 );
        return numFirstDigit+numOtherDigits+numRecursive;
    }

    int PowerBase10( unsigned int n )
    {
        int result = 1;
        for ( unsigned int i = 0; i < n; ++i )
            result *= 10;
        return result;
    }
};

三、从数字规律着手解决
这篇讲述的比较清楚
4. 解题思路

    int NumberOf1Between1AndN_Solution(int n)
    {
        if(n < 1)
            return 0;
        int sum = 0;
        int num = n;
        int base = 1;
        while(num)
        {
            int weight = num%10;
            num /= 10;
            sum += num*base;
            if(weight == 1)
                sum += n%base + 1;
            else if(weight > 1)
                sum += base;
            base *= 10;
        }
        return sum;
    }

代码时间复杂度为O(logN)