打印大于自身 N 的数字的所有排列
原文:https://www . geesforgeks . org/print-number-n 的所有排列-大于自身/
给定一个数 N ，我们的任务是打印那些大于 N 的整数 N 的排列 T3】示例:T5】
输入: N = 534 输出: 543 输入: N = 324 输出: 342、423、432
方法:为了解决这个问题，我们可以使用 C++中的next _ arrangement()方法获得 N 的所有字典上较大的排列。得到所有这些数字后，打印出来。 对于其他语言，找到数字 N 的排列打印大于 N 的数字 以下是上述方法的实现:
C++
// C++ implementation to print all the
// permutation greater than the integer N
#include <bits/stdc++.h>
using namespace std;
// Function to print all the permutation
// which are greater than N itself
void printPermutation(int N)
{
int temp = N, count = 0;
// Iterate and count the
// number of digits in N
while (temp > 0) {
count++;
temp /= 10;
}
// vector to print the
// permutations of N
vector<int> num(count);
// Store digits of N
// in the vector num
while (N > 0) {
num[count-- - 1] = N % 10;
N = N / 10;
}
// Iterate over every permutation of N
// which is greater than N
while (next_permutation(
num.begin(), num.end())) {
// Print the current permutation of N
for (int i = 0; i < num.size(); i++)
cout << num[i];
cout << "\n";
}
}
// Driver Code
int main()
{
int N = 324;
printPermutation(N);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java implementation to print all the
// permutation greater than the integer N
import java.util.*;
class GFG{
static void printPermutation(int N)
{
int temp = N, count = 0;
// Iterate and count the
// number of digits in N
while (temp > 0)
{
count++;
temp /= 10;
}
// vector to print the
// permutations of N
int[] num = new int[count];
// Store digits of N
// in the vector num
while (N > 0)
{
num[count-- - 1] = N % 10;
N = N / 10;
}
// Iterate over every permutation of N
// which is greater than N
while (next_permutation(num))
{
// Print the current permutation of N
for (int i = 0; i < num.length; i++)
System.out.print(num[i]);
System.out.print("\n");
}
}
// Function to print all the permutation
// which are greater than N itself
static boolean next_permutation(int[] p)
{
for (int a = p.length - 2; a >= 0; --a)
if (p[a] < p[a + 1])
for (int b = p.length - 1;; --b)
if (p[b] > p[a])
{
int t = p[a];
p[a] = p[b];
p[b] = t;
for (++a, b = p.length - 1; a < b; ++a, --b)
{
t = p[a];
p[a] = p[b];
p[b] = t;
}
return true;
}
return false;
}
// Driver Code
public static void main(String[] args)
{
int N = 324;
printPermutation(N);
}
}
// This code contributed by sapnasingh4991
C
// C# implementation to print all the
// permutation greater than the integer N
using System;
class GFG{
static void printPermutation(int N)
{
int temp = N, count = 0;
// Iterate and count the
// number of digits in N
while (temp > 0)
{
count++;
temp /= 10;
}
// vector to print the
// permutations of N
int[] num = new int[count];
// Store digits of N
// in the vector num
while (N > 0)
{
num[count-- - 1] = N % 10;
N = N / 10;
}
// Iterate over every permutation of N
// which is greater than N
while (next_permutation(num))
{
// Print the current permutation of N
for (int i = 0; i < num.Length; i++)
Console.Write(num[i]);
Console.Write("\n");
}
}
// Function to print all the permutation
// which are greater than N itself
static bool next_permutation(int[] p)
{
for (int a = p.Length - 2; a >= 0; --a)
if (p[a] < p[a + 1])
for (int b = p.Length - 1;; --b)
if (p[b] > p[a])
{
int t = p[a];
p[a] = p[b];
p[b] = t;
for (++a, b = p.Length - 1;
a < b; ++a, --b)
{
t = p[a];
p[a] = p[b];
p[b] = t;
}
return true;
}
return false;
}
// Driver Code
public static void Main(String[] args)
{
int N = 324;
printPermutation(N);
}
}
// This code is contributed by Rohit_ranjan
java 描述语言
<script>
// JavaScript implementation to print all the
// permutation greater than the integer N
function printPermutation(N)
{
let temp = N, count = 0;
// Iterate and count the
// number of digits in N
while (temp > 0)
{
count++;
temp = parseInt(temp / 10, 10);
}
// vector to print the
// permutations of N
let num = new Array(count);
num.fill(0);
// Store digits of N
// in the vector num
while (N > 0)
{
num[count-- - 1] = N % 10;
N = parseInt(N / 10, 10);
}
// Iterate over every permutation of N
// which is greater than N
while (next_permutation(num))
{
// Print the current permutation of N
for (let i = 0; i < num.length; i++)
document.write(num[i]);
document.write("</br>");
}
}
// Function to print all the permutation
// which are greater than N itself
function next_permutation(p)
{
for (let a = p.length - 2; a >= 0; --a)
if (p[a] < p[a + 1])
for (let b = p.length - 1;; --b)
if (p[b] > p[a])
{
let t = p[a];
p[a] = p[b];
p[b] = t;
for (++a, b = p.length - 1; a < b; ++a, --b)
{
t = p[a];
p[a] = p[b];
p[b] = t;
}
return true;
}
return false;
}
let N = 324;
printPermutation(N);
</script>
Output:
342
423
432
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