仅使用一次交换操作时可能出现的最大较小数

给定一个字符串形式的非负数N。任务是对数字N应用最多一个交换操作,使结果小于N,并且是此类数字的最大值。 例如:

null
Input :str = "12435"Output : 12345Although the number 12354 will be the largest smaller number from 12435. But it is not possible to make it using onlyone swap. So swap 4 and 3 and get 12345.Input : 34123567Output : 33124567We swap 4 with 3 (on its right side) toget the largest smaller number.Input : str = " 12345"Output : -1Digits are in increasing order. So it is not possible to make a smaller number from it.

建议:请在“上解决” 实践 “首先,在讨论解决方案之前。

  1. 从右边开始遍历,找到一个大于右边一个数字的数字。让这个索引这样的元素成为索引。
  2. 然后在str[index]右边找到另一个索引,它的最大值小于str[index]。
  3. 交换上面找到的两个值。

C++

// C++ program to find the largest smaller
// number by swapping one digit.
#include <bits/stdc++.h>
using namespace std;
// Returns largest possible number with one
// swap such that the number is smaller than
// str. It is assumed that there are leading
// 0s.
string prevNum(string str)
{
int len = str.length();
int index = -1;
// Traverse from right until we find
// a digit which is greater than its
// next digit. For example, in 34125,
// our index is 4.
for ( int i = len - 2; i >= 0; i--) {
if (str[i] > str[i+1])
{
index = i;
break ;
}
}
// We can also use binary search here as
// digits after index are sorted in increasing
// order.
// Find the biggest digit in the right of
// arr[index] which is smaller than arr[index]
int smallGreatDgt = -1;
for ( int i = len - 1; i > index; i--) {
if (str[i] < str[index]) {
if (smallGreatDgt == -1)
smallGreatDgt = i;
else if (str[i] >= str[smallGreatDgt])
smallGreatDgt = i;
}
}
// If index is -1 i.e. digits are
// in increasing order.
if (index == -1)
return "-1" ;
// Swap both values
if (smallGreatDgt != -1)
{
swap(str[index], str[smallGreatDgt]);
return str;
}
return "-1" ;
}
// Drivers code
int main()
{
string str = "34125" ;
cout << prevNum(str);
return 0;
}


JAVA

// Java program to find the largest smaller
// number by swapping one digit.
class GFG
{
// Returns largest possible number
// with one swap such that the number
// is smaller than str. It is assumed
// that there are leading 0s.
static String prevNum(String str)
{
int len = str.length();
int index = - 1 ;
// Traverse from right until we find
// a digit which is greater than its
// next digit. For example, in 34125,
// our index is 4.
for ( int i = len - 2 ; i >= 0 ; i--)
{
if (str.charAt(i) > str.charAt(i + 1 ))
{
index = i;
break ;
}
}
// We can also use binary search here as
// digits after index are sorted in increasing
// order.
// Find the biggest digit in the right of
// arr[index] which is smaller than arr[index]
int smallGreatDgt = - 1 ;
for ( int i = len - 1 ; i > index; i--)
{
if (str.charAt(i) < str.charAt(index))
{
if (smallGreatDgt == - 1 )
{
smallGreatDgt = i;
}
else if (str.charAt(i) >=
str.charAt(smallGreatDgt))
{
smallGreatDgt = i;
}
}
}
// If index is -1 i.e. digits are
// in increasing order.
if (index == - 1 )
{
return "-1" ;
}
// Swap both values
if (smallGreatDgt != - 1 )
{
str = swap(str, index, smallGreatDgt);
return str;
}
return "-1" ;
}
static String swap(String str, int i, int j)
{
char ch[] = str.toCharArray();
char temp = ch[i];
ch[i] = ch[j];
ch[j] = temp;
return String.valueOf(ch);
}
// Driver code
public static void main(String[] args)
{
String str = "34125" ;
System.out.println(prevNum(str));
}
}
// This code is contributed by 29AjayKumar


Python3

# Python3 program to find the largest smaller
# number by swapping one digit.
import sys
# Returns largest possible number
# with one swap such that the number
# is smaller than str. It is assumed
# that there are leading 0s.
def prevNum(string, n):
index = - 1
# Traverse from right until we find
# a digit which is greater than its
# next digit. For example, in 34125,
# our index is 4.
for i in range (n - 2 , - 1 , - 1 ):
if int (string[i]) > int (string[i + 1 ]):
index = i
break
# We can also use binary search here as
# digits after index are sorted in
# increasing order.
# Find the biggest digit in the right of
# arr[index] which is smaller than arr[index]
smallGreatDgt = - 1
for i in range (n - 1 , index, - 1 ):
if (smallGreatDgt = = - 1 and int (string[i]) <
int (string[index])):
smallGreatDgt = i
elif (index > - 1 and int (string[i]) > =
int (string[smallGreatDgt]) and
int (string[i]) < int (string[index])):
smallGreatDgt = i
# If index is -1 i.e. digits are
# in increasing order.
if index = = - 1 :
return " " . join(" - 1 ")
else :
# Swap both values
(string[index],
string[smallGreatDgt]) = (string[smallGreatDgt],
string[index])
return "" . join(string)
# Driver Code
if __name__ = = '__main__' :
n_str = "34125"
ans = prevNum( list (n_str), len (n_str))
print (ans)
# This code is contributed by Vikash Kumar 37


C#

// C# program to find the largest smaller
// number by swapping one digit.
using System;
class GFG
{
// Returns largest possible number
// with one swap such that the number
// is smaller than str. It is assumed
// that there are leading 0s.
static String prevNum(String str)
{
int len = str.Length;
int index = -1;
// Traverse from right until we find
// a digit which is greater than its
// next digit. For example, in 34125,
// our index is 4.
for ( int i = len - 2; i >= 0; i--)
{
if (str[i] > str[i + 1])
{
index = i;
break ;
}
}
// We can also use binary search here as
// digits after index are sorted in increasing
// order.
// Find the biggest digit in the right of
// arr[index] which is smaller than arr[index]
int smallGreatDgt = -1;
for ( int i = len - 1; i > index; i--)
{
if (str[i] < str[index])
{
if (smallGreatDgt == -1)
{
smallGreatDgt = i;
}
else if (str[i] >=
str[smallGreatDgt])
{
smallGreatDgt = i;
}
}
}
// If index is -1 i.e. digits are
// in increasing order.
if (index == -1)
{
return "-1" ;
}
// Swap both values
if (smallGreatDgt != -1)
{
str = swap(str, index, smallGreatDgt);
return str;
}
return "-1" ;
}
static String swap(String str, int i, int j)
{
char [] ch = str.ToCharArray();
char temp = ch[i];
ch[i] = ch[j];
ch[j] = temp;
return String.Join( "" ,ch);
}
// Driver code
public static void Main(String[] args)
{
String str = "34125" ;
Console.WriteLine(prevNum(str));
}
}
/* This code contributed by PrinciRaj1992 */


PHP

<?php
// PHP program to find the
// largest smaller number
// by swapping one digit.
// Returns largest possible
// number with one swap such
// that the number is smaller
// than str. It is assumed
// that there are leading
// 0s.
function prevNum( $str )
{
$len = strlen ( $str );
$index = -1;
// Traverse from right
// until we find a digit
// which is greater than
// its next digit. For
// example, in 34125,
// our index is 4.
for ( $i = $len - 2; $i >= 0; $i --)
{
if ( $str [ $i ] > $str [ $i + 1])
{
$index = $i ;
break ;
}
}
// We can also use binary
// search here as digits
// after index are sorted
// in increasing order.
// Find the biggest digit
// in the right of arr[index]
// which is smaller than
// arr[index]
$smallGreatDgt = -1;
for ( $i = $len - 1;
$i > $index ; $i --)
{
if ( $str [ $i ] < $str [ $index ])
{
if ( $smallGreatDgt == -1)
$smallGreatDgt = $i ;
else if ( $str [ $i ] >= $str [ $smallGreatDgt ])
$smallGreatDgt = $i ;
}
}
// If index is -1 i.e.
// digits are in
// increasing order.
if ( $index == -1)
return "-1" ;
// Swap both values
if ( $smallGreatDgt != -1)
{
list( $str [ $index ],
$str [ $smallGreatDgt ]) = array ( $str [ $smallGreatDgt ],
$str [ $index ]);
// swap(str[index],
// str[smallGreatDgt]);
return $str ;
}
return "-1" ;
}
// Driver code
$str = "34125" ;
echo prevNum( $str );
// This code is contributed
// by akt_mit
?>


Javascript

<script>
// Javascript program to find the largest smaller
// number by swapping one digit.
// Returns largest possible number
// with one swap such that the number
// is smaller than str. It is assumed
// that there are leading 0s.
function prevNum(str)
{
let len = str.length;
let index = -1;
// Traverse from right until we find
// a digit which is greater than its
// next digit. For example, in 34125,
// our index is 4.
for (let i = len - 2; i >= 0; i--)
{
if (str[i] > str[i + 1])
{
index = i;
break ;
}
}
// We can also use binary search here as
// digits after index are sorted in increasing
// order.
// Find the biggest digit in the right of
// arr[index] which is smaller than arr[index]
let smallGreatDgt = -1;
for (let i = len - 1; i > index; i--)
{
if (str[i] < str[index])
{
if (smallGreatDgt == -1)
{
smallGreatDgt = i;
}
else if (str[i] >= str[smallGreatDgt])
{
smallGreatDgt = i;
}
}
}
// If index is -1 i.e. digits are
// in increasing order.
if (index == -1)
{
return "-1" ;
}
// Swap both values
if (smallGreatDgt != -1)
{
str = swap(str, index, smallGreatDgt);
return str;
}
return "-1" ;
}
function swap(str, i, j)
{
let ch = str.split( '' );
let temp = ch[i];
ch[i] = ch[j];
ch[j] = temp;
return ch.join( "" );
}
let str = "34125" ;
document.write(prevNum(str));
</script>


输出:

32145

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