找到最多换 K 位可以组成的最大数字
原文:https://www . geeksforgeeks . org/find-最多改变 k 位数可以形成的最大数字/
给定代表一个数的字符串 str 和一个整数 K ，任务是找出给定数中最多改变 K 位可以构成的最大数。
示例:
输入: str = "569431 "，K = 3 输出: 999931 将第一、二、四位数字替换为 9。 输入: str = "5687 "，K = 2 输出: 9987
方法:为了得到可能的最大数字，最左边的数字必须替换为 9s。对于从最左边的数字开始的数字的每一个数字，如果它还不是 9，并且 K 大于 0，那么用 9 替换它，并将 K 减 1。当 K 大于 0 时，对每个数字重复这些步骤。最后，打印更新后的号码。
下面是上述方法的实现:
C++
// C++ implementation of the approach
#include <iostream>
using namespace std;
// Function to return the maximum number
// that can be formed by changing
// at most k digits in str
string findMaximumNum(string str, int n, int k)
{
// For every digit of the number
for (int i = 0; i < n; i++) {
// If no more digits can be replaced
if (k < 1)
break;
// If current digit is not already 9
if (str[i] != '9') {
// Replace it with 9
str[i] = '9';
// One digit has been used
k--;
}
}
return str;
}
// Driver code
int main()
{
string str = "569431";
int n = str.length();
int k = 3;
cout << findMaximumNum(str, n, k);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java implementation of the approach
class GFG
{
// Function to return the maximum number
// that can be formed by changing
// at most k digits in str
static StringBuilder findMaximumNum(StringBuilder str,
int n, int k)
{
// For every digit of the number
for (int i = 0; i < n; i++)
{
// If no more digits can be replaced
if (k < 1)
break;
// If current digit is not already 9
if (str.charAt(i) != '9')
{
// Replace it with 9
str.setCharAt(i, '9');
// One digit has been used
k--;
}
}
return str;
}
// Driver code
public static void main (String [] args)
{
StringBuilder str = new StringBuilder("569431");
int n = str.length();
int k = 3;
System.out.println(findMaximumNum(str, n, k));
}
}
// This code is contributed by ihritik
Python 3
# Python3 implementation of the approach
# Function to return the maximum number
# that can be formed by changing
# at most k digits in str
def findMaximumNum(st, n, k):
# For every digit of the number
for i in range(n):
# If no more digits can be replaced
if (k < 1):
break
# If current digit is not already 9
if (st[i] != '9'):
# Replace it with 9
st = st[0:i] + '9' + st[i + 1:]
# One digit has been used
k -= 1
return st
# Driver code
st = "569431"
n = len(st)
k = 3
print(findMaximumNum(st, n, k))
# This code is contributed by
# divyamohan123
C
// C# implementation of the approach
using System;
using System.Text;
class GFG
{
// Function to return the maximum number
// that can be formed by changing
// at most k digits in str
static StringBuilder findMaximumNum(StringBuilder str,
int n, int k)
{
// For every digit of the number
for (int i = 0; i < n; i++)
{
// If no more digits can be replaced
if (k < 1)
break;
// If current digit is not already 9
if (str[i] != '9')
{
// Replace it with 9
str[i] = '9';
// One digit has been used
k--;
}
}
return str;
}
// Driver code
public static void Main ()
{
StringBuilder str = new StringBuilder("569431");
int n = str.Length;
int k = 3;
Console.WriteLine(findMaximumNum(str, n, k));
}
}
// This code is contributed by ihritik
java 描述语言
<script>
// JavaScript implementation of the approach
// Function to return the maximum number
// that can be formed by changing
// at most k digits in str
function findMaximumNum(str, n, k) {
// For every digit of the number
for (var i = 0; i < n; i++) {
// If no more digits can be replaced
if (k < 1) break;
// If current digit is not already 9
if (str[i] !== "9") {
// Replace it with 9
str[i] = "9";
// One digit has been used
k--;
}
}
return str.join("");
}
// Driver code
var str = "569431";
var n = str.length;
var k = 3;
document.write(findMaximumNum(str.split(""), n, k));
</script>
Output:
999931
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