给定N个非负整数的数组arr [] ,任务是根据这些整数的和对这些整数进行排序。
例子:
Input: arr[] = {12, 10, 102, 31, 15}
Output: 10 12 102 31 15
10 => 1 + 0 = 1
12 => 1 + 2 = 3
102 => 1 + 0 + 2 = 3
31 => 3 + 1= 4
15 => 1 + 5 = 6
Input: arr[] = {14, 1101, 10, 35, 0}
Output: 0 10 1101 14 35
方法:想法是将每个元素及其位数之和存储在向量对中,然后根据所存储的位数之和对向量的所有元素进行排序。最后,按顺序打印元素。
下面是上述方法的实现:
C++
// C++ implementation of the approach
#include
using namespace std;
// Function to return the sum
// of the digits of n
int sumOfDigit(int n)
{
int sum = 0;
while (n > 0) {
sum += n % 10;
n = n / 10;
}
return sum;
}
// Function to sort the array according to
// the sum of the digits of elements
void sortArr(int arr[], int n)
{
// Vector to store digit sum with respective element
vector > vp;
// Inserting digit sum with element in the vector pair
for (int i = 0; i < n; i++) {
vp.push_back(make_pair(sumOfDigit(arr[i]), arr[i]));
}
// Quick sort the vector, this will sort the pair
// according to sum of the digits of elements
sort(vp.begin(), vp.end());
// Print the sorted vector content
for (int i = 0; i < vp.size(); i++)
cout << vp[i].second << " ";
}
// Driver code
int main()
{
int arr[] = { 14, 1101, 10, 35, 0 };
int n = sizeof(arr) / sizeof(arr[0]);
sortArr(arr, n);
return 0;
}
Python3
# Python3 implementation of the approach
# Function to return the sum
# of the digits of n
def sumOfDigit(n) :
sum = 0;
while (n > 0) :
sum += n % 10;
n = n // 10;
return sum;
# Function to sort the array according to
# the sum of the digits of elements
def sortArr(arr, n) :
# Vector to store digit sum with
# respective element
vp = [];
# Inserting digit sum with element
# in the vector pair
for i in range(n) :
vp.append([sumOfDigit(arr[i]), arr[i]]);
# Quick sort the vector, this will
# sort the pair according to sum
# of the digits of elements
vp.sort()
# Print the sorted vector content
for i in range(len(vp)) :
print(vp[i][1], end = " ");
# Driver code
if __name__ == "__main__" :
arr = [14, 1101, 10, 35, 0];
n = len(arr);
sortArr(arr, n);
# This code is contributed by Ryuga
输出:
0 10 1101 14 35