给定一个以数组形式表示的数字，使得数组的每个元素存储该数字的一位数。也就是说，数字 1234 的数组将是 arr[] = {1,2,3,4}。任务是找到比给定数字大但数字总和等于给定数字的最小数字。
为简单起见：考虑数字的长度最多可以是 20。
例子：
Input : arr[] = {0, 0, 0, 0, 0, 0, 0, 3, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 8 };
Output : 00000004799999999999
Explanation : Sum of digits = 110
Input : arr[] = {0, 0, 0, 0, 0, 0, 0, 3, 9, 7, 0, 0, 2, 9, 8, 9, 5, 9, 9, 0};
Output : 00000003970029896089
Explanation : Sum of digits = 70
蛮力方法是：
- 从该数字开始并将数字递增 1。
- 检查总和。如果总和相同，则返回数字。
- 否则返回第一步。
更好的方法是跳到 O(n) 时间复杂度中的下一个数字，其中 n 是字符串的长度。
We divide the number into 4 regions :
1st: Trailing zeros .
2nd: The lowest digit not in Region 1.
3rd: Consecutive 9s starting with the digit above Region 2.
4th: All remaining digits.
Then the next number is :
[Region 4+1] [Region 1] [Region 2-1] [Region 3] .
Example:
Input Number = 548995000
Region 1 : 000
Region 2 : 5
Region 3 : 99
Region 4 : 548
Next number = 549000499
下面是上述方法的实现：
C++
// CPP program to find next greater number with
// same sum of digits.
#include
using namespace std;
#define pb push_back
void getnext(int arr[], int n)
{
// for storing 4 regions
vector a1, a2, a3, a4;
// trailing zeros region1
int i = n - 1; // last index
while (arr[i] == 0)
{
a1.pb(0);
i--;
}
// lowest region(region 2) not in region 1
a2.pb(arr[i--]);
// Consecutive 9's (region 3)
while (arr[i] == 9)
{
a3.pb(9);
i--;
}
int j = 0;
while (arr[j] == 0)
j++; // Starting zeros
while (j <= i) // 4th region
{
a4.pb(arr[j]);
j++;
}
// Calculation of result
j = a4.size() - 1;
a4[j]++; // Region4 + 1
a2[0]--; // Region2 -1
int l = a1.size() + a2.size() + a3.size() + a4.size();
// Calculating the result
j = n-1;
i = a3.size() - 1;
// Copying the third region
while (i >= 0)
{
arr[j--] = a3[i--];
}
// Copying the 2nd region
i = a2.size() - 1;
while (i >= 0)
{
arr[j--] = a2[i--];
}
// Copying the 1st region
i = a1.size() - 1;
while (i >= 0)
{
arr[j--] = a1[i--];
}
// Copying the 4th region
i = a4.size() - 1;
while (i >= 0)
{
arr[j--] = a4[i--];
}
while (j >= 0)
arr[j--] = 0;
}
int main()
{
int arr[] = { 0, 0, 0, 0, 0, 0, 0, 3, 9, 7,
0, 0, 2, 9, 8, 9, 5, 9, 9, 0 };
int n = sizeof(arr)/sizeof(arr[0]);
getnext(arr, n); // Calling the function
for (int i = 0; i < n; i++)
cout << arr[i];
return 0;
} Java
// Java program to find next greater number with
// same sum of digits.
import java.util.*;
class GFG
{
static void getnext(int []arr, int n)
{
// for storing 4 regions
ArrayList a1 = new ArrayList();
ArrayList a2 = new ArrayList();
ArrayList a3 = new ArrayList();
ArrayList a4 = new ArrayList();
// trailing zeros region1
int i = n - 1; // last index
while (arr[i] == 0)
{
a1.add(0);
i--;
}
// lowest region(region 2) not in region 1
a2.add(arr[i--]);
// Consecutive 9's (region 3)
while (arr[i] == 9)
{
a3.add(9);
i--;
}
int j = 0;
while (arr[j] == 0)
j++; // Starting zeros
while (j <= i) // 4th region
{
a4.add(arr[j]);
j++;
}
// Calculation of result
j = a4.size() - 1;
a4.set(j,a4.get(j) + 1); // Region4 + 1
a2.set(0,a2.get(0) - 1); // Region2 -1
//int l = a1.size() + a2.size() + a3.size() + a4.size();
// Calculating the result
j = n - 1;
i = a3.size() - 1;
// Copying the third region
while (i >= 0)
{
arr[j--] = (int)a3.get(i);
i--;
}
// Copying the 2nd region
i = a2.size() - 1;
while (i >= 0)
{
arr[j--] = (int)a2.get(i);
i--;
}
// Copying the 1st region
i = a1.size() - 1;
while (i >= 0)
{
arr[j--] = a1.get(i);
i--;
}
// Copying the 4th region
i = a4.size() - 1;
while (i >= 0)
{
arr[j--] = a4.get(i);
i--;
}
while (j >= 0)
arr[j--] = 0;
}
// Driver code
public static void main (String[] args)
{
int []arr = { 0, 0, 0, 0, 0, 0, 0, 3, 9, 7,
0, 0, 2, 9, 8, 9, 5, 9, 9, 0 };
int n = arr.length;
getnext(arr, n); // Calling the function
for (int i = 0; i < n; i++)
System.out.print(arr[i]);
}
}
// This code is contributed by mits Python3
# Python3 program to find next greater number with
# same sum of digits.
def getnext(arr, n):
# for storing 4 regions
a1=[];
a2=[];
a3=[];
a4=[];
# trailing zeros region1
i = n - 1; # last index
while (arr[i] == 0):
a1.append(0);
i-=1;
# lowest region(region 2) not in region 1
a2.append(arr[i]);
i-=1;
# Consecutive 9's (region 3)
while (arr[i] == 9):
a3.append(9);
i-=1;
j = 0;
while (arr[j] == 0):
j+=1; # Starting zeros
while (j <= i): # 4th region
a4.append(arr[j]);
j+=1;
# Calculation of result
j = len(a4) - 1;
a4[j]+=1; # Region4 + 1
a2[0]-=1; # Region2 -1
l = len(a1) + len(a2) + len(a3) + len(a4);
# Calculating the result
j = n-1;
i = len(a3) - 1;
# Copying the third region
while (i >= 0):
arr[j] = a3[i];
j-=1;
i-=1;
# Copying the 2nd region
i = len(a2) - 1;
while (i >= 0):
arr[j] = a2[i];
j-=1;
i-=1;
# Copying the 1st region
i = len(a1) - 1;
while (i >= 0):
arr[j] = a1[i];
j-=1;
i-=1;
# Copying the 4th region
i = len(a4) - 1;
while (i >= 0):
arr[j] = a4[i];
j-=1;
i-=1;
while (j >= 0):
arr[j] = 0;
j-=1;
# Driver code
arr = [ 0, 0, 0, 0, 0, 0, 0, 3, 9, 7, 0,
0, 2, 9, 8, 9, 5, 9, 9, 0 ];
n = len(arr);
getnext(arr, n); # Calling the function
for i in range(0,n):
print(arr[i],end="");
# This code is contributed by mitsC#
// C# program to find next greater number with
// same sum of digits.
using System;
using System.Collections;
class GFG
{
static void getnext(int []arr, int n)
{
// for storing 4 regions
ArrayList a1 = new ArrayList();
ArrayList a2 = new ArrayList();
ArrayList a3 = new ArrayList();
ArrayList a4 = new ArrayList();
// trailing zeros region1
int i = n - 1; // last index
while (arr[i] == 0)
{
a1.Add(0);
i--;
}
// lowest region(region 2) not in region 1
a2.Add(arr[i--]);
// Consecutive 9's (region 3)
while (arr[i] == 9)
{
a3.Add(9);
i--;
}
int j = 0;
while (arr[j] == 0)
j++; // Starting zeros
while (j <= i) // 4th region
{
a4.Add(arr[j]);
j++;
}
// Calculation of result
j = a4.Count - 1;
a4[j] = (int)a4[j] + 1; // Region4 + 1
a2[0] = (int)a2[0] - 1; // Region2 -1
//int l = a1.Count + a2.Count + a3.Count + a4.Count;
// Calculating the result
j = n - 1;
i = a3.Count - 1;
// Copying the third region
while (i >= 0)
{
arr[j--] = (int)a3[i];
i--;
}
// Copying the 2nd region
i = a2.Count - 1;
while (i >= 0)
{
arr[j--] = (int)a2[i];
i--;
}
// Copying the 1st region
i = a1.Count - 1;
while (i >= 0)
{
arr[j--] = (int)a1[i];
i--;
}
// Copying the 4th region
i = a4.Count - 1;
while (i >= 0)
{
arr[j--] = (int)a4[i];
i--;
}
while (j >= 0)
arr[j--] = 0;
}
// Driver code
static void Main()
{
int []arr = { 0, 0, 0, 0, 0, 0, 0, 3, 9, 7,
0, 0, 2, 9, 8, 9, 5, 9, 9, 0 };
int n = arr.Length;
getnext(arr, n); // Calling the function
for (int i = 0; i < n; i++)
Console.Write(arr[i]);
}
}
// This code is contributed by mitsPHP
= 0)
{
$arr[$j--] = $a3[$i--];
}
// Copying the 2nd region
$i = count($a2) - 1;
while ($i >= 0)
{
$arr[$j--] = $a2[$i--];
}
// Copying the 1st region
$i = count($a1) - 1;
while ($i >= 0)
{
$arr[$j--] = $a1[$i--];
}
// Copying the 4th region
$i = count($a4) - 1;
while ($i >= 0)
{
$arr[$j--] = $a4[$i--];
}
while ($j >= 0)
$arr[$j--] = 0;
}
// Driver code
$arr = array( 0, 0, 0, 0, 0, 0, 0, 3, 9, 7,
0, 0, 2, 9, 8, 9, 5, 9, 9, 0 );
$n = count($arr);
getnext($arr, $n); // Calling the function
for ($i = 0; $i < $n; $i++)
echo $arr[$i];
// This code is contributed by mits
?>输出：
00000003970029896089
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