给定数字n，请找到一个介于1到n之间的数字，以使其总和最大。如果有几个这样的整数，请确定其中最大的一个。
例子 ：
Input: n = 100
Output: 99
99 is the largest number in range from
1 to 100 with maximum sum of digits.
Input: n = 48
Output: 48
Explanation:
There are two numbers with maximum
digit sum. The numbers are 48 and 39
Since 48 > 39, it is our answer.
天真的方法是对从1到n的所有数字进行迭代，并找出哪个数字的位数之和最大。该解决方案的时间复杂度为O(n)。
一种有效的方法是从n迭代到1。对当前数字的每个数字执行以下操作，如果数字不为零，则将其减少一位，然后将所有其他数字更改为右边的九个数字。如果结果整数中的位数之和严格大于当前答案的位数之和，则使用结果整数更新答案。如果结果整数的总和与当前答案相同，则如果结果整数大于当前答案，则用结果整数更新当前答案。
如何减少一个数字并将其右边的所有其他数字更改为9?
令x为当前数字。我们可以使用以下公式找到当前数字的下一个数字。在下面的公式中，b是10的幂代表当前数字的位置。每次迭代后，我们将x减小为x / 10并将b更改为b * 10。
我们使用(x – 1)* b +(b – 1);
此行可以进一步解释为，如果数字为x = 521且b = 1，则
- (521 – 1)* 1 +(1-1)为您提供520，这是我们需要做的事情，将位置编号减少1，并将所有其他右侧的位置编号替换为9。
- 在x / = 10给出x为52并且b * = 10给出b为10之后，再次执行(52-1)*(10)+ 9得到519，这是我们要做的，将当前索引减少1，并将所有其他权利增加9。
C++
// CPP program to find the
// number with maximum digit
// sum.
#include
using namespace std;
// function to calculate the
// sum of digits of a number.
int sumOfDigits(int a)
{
int sum = 0;
while (a)
{
sum += a % 10;
a /= 10;
}
return sum;
}
// Returns the maximum number
// with maximum sum of digits.
int findMax(int x)
{
// initializing b as 1 and
// initial max sum to be of n
int b = 1, ans = x;
// iterates from right to
// left in a digit
while (x)
{
// while iterating this
// is the number from
// from right to left
int cur = (x - 1) * b + (b - 1);
// calls the function to
// check if sum of cur is
// more then of ans
if (sumOfDigits(cur) > sumOfDigits(ans) ||
(sumOfDigits(cur) == sumOfDigits(ans) &&
cur > ans))
ans = cur;
// reduces the number to one unit less
x /= 10;
b *= 10;
}
return ans;
}
// driver program
int main()
{
int n = 521;
cout << findMax(n);
return 0;
} Java
// Java program to find the
// number with maximum digit
// sum.
import java.io.*;
class GFG {
// function to calculate the
// sum of digits of a number.
static int sumOfDigits(int a)
{
int sum = 0;
while (a!=0)
{
sum += a % 10;
a /= 10;
}
return sum;
}
// Returns the maximum number
// with maximum sum of digits.
static int findMax(int x)
{
// initializing b as 1 and
// initial max sum to be of n
int b = 1, ans = x;
// iterates from right to
// left in a digit
while (x!=0)
{
// while iterating this
// is the number from
// from right to left
int cur = (x - 1) * b + (b - 1);
// calls the function to
// check if sum of cur is
// more then of ans
if (sumOfDigits(cur) > sumOfDigits(ans) ||
(sumOfDigits(cur) == sumOfDigits(ans) &&
cur > ans))
ans = cur;
// reduces the number to one unit less
x /= 10;
b *= 10;
}
return ans;
}
// driver program
public static void main (String[] args)
{
int n = 521;
System.out.println(findMax(n));
}
}
/*This article is contributed by Nikita Tiwari.*/Python3
# Python 3 program to
# find the number
# with maximum digit
# sum.
# function to calculate
# the sum of digits of
# a number.
def sumOfDigits(a) :
sm = 0
while (a!=0) :
sm = sm + a % 10
a = a // 10
return sm
# Returns the maximum number
# with maximum sum of digits.
def findMax(x) :
# initializing b as 1
# and initial max sum
# to be of n
b = 1
ans = x
# iterates from right
# to left in a digit
while (x!=0) :
# while iterating this
# is the number from
# right to left
cur = (x - 1) * b + (b - 1)
# calls the function to
# check if sum of cur is
# more then of ans
if (sumOfDigits(cur) > sumOfDigits(ans) or
(sumOfDigits(cur) == sumOfDigits(ans) and
cur > ans)) :
ans = cur
# reduces the number
# to one unit less
x =x // 10
b = b * 10
return ans
# driver program to test the above function
n = 521
print(findMax(n))
# This article is contributed by Nikita Tiwari.C#
// C# program to find the number
// with maximum digit sum.
using System;
class GFG {
// function to calculate the
// sum of digits of a number.
static int sumOfDigits(int a)
{
int sum = 0;
while (a!=0)
{
sum += a % 10;
a /= 10;
}
return sum;
}
// Returns the maximum number
// with maximum sum of digits.
static int findMax(int x)
{
// initializing b as 1 and
// initial max sum to be of n
int b = 1, ans = x;
// iterates from right to
// left in a digit
while (x!=0)
{
// while iterating this
// is the number from
// from right to left
int cur = (x - 1) * b + (b - 1);
// calls the function to
// check if sum of cur is
// more then of ans
if (sumOfDigits(cur) > sumOfDigits(ans) ||
(sumOfDigits(cur) == sumOfDigits(ans) &&
cur > ans))
ans = cur;
// reduces the number to one unit less
x /= 10;
b *= 10;
}
return ans;
}
// driver program
public static void Main()
{
int n = 521;
Console.WriteLine(findMax(n));
}
}
// This article is contributed by Anant Agarwal.PHP
sumOfDigits($ans) ||
(sumOfDigits($cur) == sumOfDigits($ans) &&
$cur > $ans))
$ans = $cur;
// reduces the number
// to one unit less
$x = (int)$x / 10;
$b *= 10;
}
return $ans;
}
// Driver Code
$n = 521;
echo findMax($n);
// This code is contributed by ajit
?>输出 ：
499
时间复杂度： O(m)其中m是n中的位数。