通过互换数字
最接近较大的数字
给定两个整数 A 和 B 。任务是通过互换 A 的数字找到与 B 最近的较大值。如果没有这样的可能，那么打印-1。 举例:
输入: A = 459，B = 500 输出: 549 549 是最近的较大者。 输入: A = 321，B = 567 输出: -1 输入: A = 231，B = 125 输出: 132
先决条件: 一串的所有排列 方法:
- 使用整数设置 min1 的最小值。最大值
- 用上述置换方法交换 A 的数字。
- 检查 A 的介电常数是否小于 min1。如果小于，则将 min1 更新为 a。
- 对 A 的所有置换重复此操作，找到最小的较大值
以下是上述方法的实现:
C++
// C++ program to find nearest greater value
#include <bits/stdc++.h>
using namespace std;
int min1 = INT_MAX;
int _count = 0;
// Find all the possible permutation of Value A.
int permutation(string str1, int i, int n, int p)
{
if (i == n)
{
// Convert into Integer
int q = stoi(str1);
// Find the minimum value of A by interchanging
// the digit of A and store min1.
if (q - p > 0 && q < min1)
{
min1 = q;
_count = 1;
}
}
else
{
for (int j = i; j <= n; j++)
{
// Swap two string character
swap(str1[i], str1[j]);
permutation(str1, i + 1, n, p);
swap(str1[i], str1[j]);
}
}
return min1;
}
// Driver code
int main()
{
int A = 213;
int B = 111;
// Convert integer value into string to
// find all the permutation of the number
string str1 = to_string(A);
int len = str1.length();
int h = permutation(str1, 0, len - 1, B);
// count=1 means number greater than B exists
_count ? cout << h << endl : cout << -1 << endl;
return 0;
}
// This code is contributed by
// sanjeev2552
Java 语言(一种计算机语言，尤用于创建网站)
// JAVA program to find nearest greater value
import java.io.*;
import java.util.*;
class GFG {
static int min1 = Integer.MAX_VALUE;
static int count = 0;
// Find all the possible permutation of Value A.
public int permutation(String str1, int i, int n, int p)
{
if (i == n) {
// Convert into Integer
int q = Integer.parseInt(str1);
// Find the minimum value of A by interchanging
// the digit of A and store min1.
if (q - p > 0 && q < min1) {
min1 = q;
count = 1;
}
}
else {
for (int j = i; j <= n; j++) {
str1 = swap(str1, i, j);
permutation(str1, i + 1, n, p);
str1 = swap(str1, i, j);
}
}
return min1;
}
// Swap two string character
public String swap(String str, int i, int j)
{
char ch[] = str.toCharArray();
char temp = ch[i];
ch[i] = ch[j];
ch[j] = temp;
// Return the string after
// swapping between two character.
return String.valueOf(ch);
}
// Driver code
public static void main(String[] args)
{
int A = 213;
int B = 111;
GFG gfg = new GFG();
// Convert integer value into string to
// find all the permutation of the number
String str1 = Integer.toString(A);
int len = str1.length();
int h = gfg.permutation(str1, 0, len - 1, B);
// count=1 means number greater than B exists
if (count == 1)
System.out.println(h);
else
System.out.println(-1);
}
}
Python 3
# Python3 program to find nearest greater value
min1 = 10**9
_count = 0
# Find all the possible permutation of Value A.
def permutation(str1, i, n, p):
global min1, _count
if (i == n):
# Convert into Integer
str1 = "".join(str1)
q = int(str1)
# Find the minimum value of A
# by interchanging the digits
# of A and store min1.
if (q - p > 0 and q < min1):
min1 = q
_count = 1
else:
for j in range(i, n + 1):
# Swap two character)
str1[i], str1[j] = str1[j], str1[i]
permutation(str1, i + 1, n, p)
str1[i], str1[j] = str1[j], str1[i]
return min1
# Driver code
A = 213
B = 111
# Convert integer value into to
# find all the permutation of the number
str2 = str(A)
str1 = [i for i in str2]
le = len(str1)
h = permutation(str1, 0, le - 1, B)
# count=1 means number greater than B exists
if _count == 1:
print(h)
else:
print(-1)
# This code is contributed by
# mohit
C
// C# program to find nearest greater value
using System;
class GFG
{
static int min1 = int.MaxValue;
static int count = 0;
// Find all the possible permutation of Value A.
public int permutation(String str1, int i,
int n, int p)
{
if (i == n)
{
// Convert into Integer
int q = int.Parse(str1);
// Find the minimum value of A by interchanging
// the digit of A and store min1.
if (q - p > 0 && q < min1)
{
min1 = q;
count = 1;
}
}
else
{
for (int j = i; j <= n; j++)
{
str1 = swap(str1, i, j);
permutation(str1, i + 1, n, p);
str1 = swap(str1, i, j);
}
}
return min1;
}
// Swap two string character
public String swap(String str, int i, int j)
{
char []ch = str.ToCharArray();
char temp = ch[i];
ch[i] = ch[j];
ch[j] = temp;
// Return the string after
// swapping between two character.
return String.Join("", ch);
}
// Driver code
public static void Main(String[] args)
{
int A = 213;
int B = 111;
GFG gfg = new GFG();
// Convert integer value into string to
// find all the permutation of the number
String str1 = A.ToString();
int len = str1.Length;
int h = gfg.permutation(str1, 0, len - 1, B);
// count=1 means number greater than B exists
if (count == 1)
Console.WriteLine(h);
else
Console.WriteLine(-1);
}
}
// This code is contributed by PrinciRaj1992
java 描述语言
<script>
// JAVAscript program to find nearest greater value
let min1 = Number.MAX_VALUE;
let count = 0;
// Find all the possible permutation of Value A.
function permutation(str1,i,n,p)
{
if (i == n) {
// Convert into Integer
let q = parseInt(str1);
// Find the minimum value of A by interchanging
// the digit of A and store min1.
if (q - p > 0 && q < min1) {
min1 = q;
count = 1;
}
}
else {
for (let j = i; j <= n; j++) {
str1 = swap(str1, i, j);
permutation(str1, i + 1, n, p);
str1 = swap(str1, i, j);
}
}
return min1;
}
// Swap two string character
function swap(str,i,j)
{
let ch = str.split("");
let temp = ch[i];
ch[i] = ch[j];
ch[j] = temp;
// Return the string after
// swapping between two character.
return (ch).join("");
}
// Driver code
let A = 213;
let B = 111;
// Convert integer value into string to
// find all the permutation of the number
let str1 = A.toString();
let len = str1.length;
let h = permutation(str1, 0, len - 1, B);
// count=1 means number greater than B exists
if (count == 1)
document.write(h);
else
document.write(-1);
// This code is contributed by unknown2108
</script>
Output:
123
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