输入整数中的所有“ 0”替换为“ 5”

给定一个整数作为输入，并将整数中的所有“ 0”替换为“ 5”。

例子：

Input: 102 
Output: 152
Explantion: All the digits which are '0' is replaced by '5' 

Input: 1020 
Output: 1525
Explantion: All the digits which are '0' is replaced by '5'

不允许使用数组存储所有数字。
资料来源：亚马逊采访体验|设置136(对于SDE-T)

迭代方法：通过观察测试用例，可以清楚地看到所有0位都由5代替。例如，对于input = 1020，output = 1525，可以写成1020 + 505，可以进一步写成1020 + 5 *(10 ^ 2)+ 5 *(10 ^ 0)。因此，可以以迭代方式形成解决方案，如果遇到“ 0”数字，则找到该数字的位置值，然后将其乘以5，然后找到数字中所有0的总和。将该总和加到输入数字中以找到输出数字。

算法：

1. 创建一个变量sum = 0以存储总和， place = 1以存储当前数字的位值，并创建输入变量的副本
2. 如果数字为零，则返回5
3. 输入变量大于0时，迭代下一步
4. 提取最后一位(n％10) ，如果该位为零，则更新sum = sum + place * 5 ，从数字n = n / 10中删除最后一位，并更新place = place * 10
5. 返回总和。

C++

#include
using namespace std;
 
// Returns the number to be added to the
// input to replace all zeroes with five
int calculateAddedValue(int number)
{
     
    // Amount to be added
    int result = 0;
 
    // Unit decimal place
    int decimalPlace = 1;
     
    if (number == 0)
    {
        result += (5 * decimalPlace);
    }
 
    while (number > 0)
    {
        if (number % 10 == 0)
        {
             
            // A number divisible by 10, then
            // this is a zero occurrence in
            // the input
            result += (5 * decimalPlace);
 
        }
         
        // Move one decimal place
        number /= 10;
        decimalPlace *= 10;
    }
    return result;
}
 
int replace0with5(int number)
{
    return number += calculateAddedValue(number);
}
 
// Driver code
int main()
{
    cout << replace0with5(1020);
}
 
// This code is contributed by avanitrachhadiya2155

Java

public class ReplaceDigits {
    static int replace0with5(int number)
    {
        return number += calculateAddedValue(number);
    }
 
    // returns the number to be added to the
    // input to replace all zeroes with five
    private static int calculateAddedValue(int number)
    {
 
        // amount to be added
        int result = 0;
 
        // unit decimal place
        int decimalPlace = 1;
 
        if (number == 0) {
            result += (5 * decimalPlace);
        }
 
        while (number > 0) {
            if (number % 10 == 0)
                // a number divisible by 10, then
                // this is a zero occurrence in the input
                result += (5 * decimalPlace);
 
            // move one decimal place
            number /= 10;
            decimalPlace *= 10;
        }
        return result;
    }
 
    public static void main(String[] args)
    {
        System.out.print(replace0with5(1020));
    }
}

Python3

def replace0with5(number):
    number += calculateAddedValue(number)
    return number
     
# returns the number to be added to the
# input to replace all zeroes with five
def calculateAddedValue(number):
     
    # amount to be added
    result = 0
     
    # unit decimal place
    decimalPlace = 1
 
    if (number == 0):
        result += (5 * decimalPlace)
         
    while (number > 0):
        if (number % 10 == 0):
             
            # a number divisible by 10, then
            # this is a zero occurrence in the input
            result += (5 * decimalPlace)
             
        # move one decimal place
        number //= 10
        decimalPlace *= 10
     
    return result
     
# Driver code
print(replace0with5(1020))
     
# This code is contributed by shubhmasingh10

C#

using System;
 
class GFG{
     
static int replace0with5(int number)
{
    return number += calculateAddedValue(number);
}
 
// Returns the number to be added to the
// input to replace all zeroes with five
static int calculateAddedValue(int number)
{
     
    // Amount to be added
    int result = 0;
 
    // Unit decimal place
    int decimalPlace = 1;
 
    if (number == 0)
    {
        result += (5 * decimalPlace);
    }
 
    while (number > 0)
    {
        if (number % 10 == 0)
         
            // A number divisible by 10, then
            // this is a zero occurrence in the input
            result += (5 * decimalPlace);
 
        // Move one decimal place
        number /= 10;
        decimalPlace *= 10;
    }
    return result;
}
 
// Driver Code
static public void Main()
{
    Console.WriteLine(replace0with5(1020));
}
}
 
// This code is contributed by rag2127

Javascript

 
// Returns the number to be added to the
// input to replace all zeroes with five
function calculateAddedValue(number){
    // Amount to be added
    let result = 0;
    // Unit decimal place
    let decimalPlace = 1;
     
    if (number == 0) {
        result += (5 * decimalPlace);
    }
 
    while (number > 0){
        if (number % 10 == 0){
            // A number divisible by 10, then
            // this is a zero occurrence in
            // the input
            result += (5 * decimalPlace);
        }
        // Move one decimal place
        number = Math.floor(number/10);
        decimalPlace *= 10;
    }
    return result;
}
 
function replace0with5(number){
    return number += calculateAddedValue(number);
}
 
// Driver code
document.write(replace0with5(1020));
 

输出：

1525

复杂度分析：

• 时间复杂度： O(k)，循环仅运行k次，其中k是该数字的位数。
• 空间复杂度： O(1)，不需要额外的空间。

递归方法：这个想法很简单，我们使用mod运算符’％’获得最后一位数字。如果数字为0，我们将其替换为5，否则，请保持原样。然后，我们再次查询剩余的数字。方法保持不变，基本区别是循环被递归函数代替。

算法：

1. 当数字为0时，请检查一个基本情况是否返回5，所有其他情况都构成一个递归函数。
2. 函数( solve(int n) )可以定义如下，如果传递的数字为0，则返回0，否则提取最后一位，即n = n / 10并删除最后一位。如果最后一位数字为零，则为其分配5。
3. 现在，通过为n调用递归函数来返回值，即返回resolve(n)* 10 + digit 。
4. 打印答案。

C++

// C++ program to replace all ‘0’
// with ‘5’ in an input Integer
#include 
using namespace std;
 
// A recursive function to replace all 0s
// with 5s in an input number It doesn't
// work if input number itself is 0.
int convert0To5Rec(int num)
{
    // Base case for recursion termination
    if (num == 0)
        return 0;
 
    // Extraxt the last digit and
    // change it if needed
    int digit = num % 10;
    if (digit == 0)
        digit = 5;
 
    // Convert remaining digits and
    // append the last digit
    return convert0To5Rec(num / 10) * 10 + digit;
}
 
// It handles 0 and calls convert0To5Rec()
// for other numbers
int convert0To5(int num)
{
    if (num == 0)
        return 5;
    else
        return convert0To5Rec(num);
}
 
// Driver code
int main()
{
    int num = 10120;
    cout << convert0To5(num);
    return 0;
}
 
// This code is contributed by Code_Mech.

C

// C program to replace all ‘0’
// with ‘5’ in an input Integer
#include 
 
// A recursive function to replace
// all 0s with 5s in an input number
// It doesn't work if input number itself is 0.
int convert0To5Rec(int num)
{
    // Base case for recursion termination
    if (num == 0)
        return 0;
 
    // Extract the last digit and change it if needed
    int digit = num % 10;
    if (digit == 0)
        digit = 5;
 
    // Convert remaining digits
    // and append the last digit
    return convert0To5Rec(num / 10) * 10 + digit;
}
 
// It handles 0 and calls
// convert0To5Rec() for other numbers
int convert0To5(int num)
{
    if (num == 0)
        return 5;
    else
        return convert0To5Rec(num);
}
 
// Driver program to test above function
int main()
{
    int num = 10120;
    printf("%d", convert0To5(num));
    return 0;
}

Java

// Java code for Replace all 0 with
// 5 in an input Integer
class GFG {
 
    // A recursive function to replace all 0s with 5s in
    // an input number. It doesn't work if input number
    // itself is 0.
    static int convert0To5Rec(int num)
    {
        // Base case
        if (num == 0)
            return 0;
 
        // Extraxt the last digit and change it if needed
        int digit = num % 10;
        if (digit == 0)
            digit = 5;
 
        // Convert remaining digits and append the
        // last digit
        return convert0To5Rec(num / 10) * 10 + digit;
    }
 
    // It handles 0 and calls convert0To5Rec() for
    // other numbers
    static int convert0To5(int num)
    {
        if (num == 0)
            return 5;
        else
            return convert0To5Rec(num);
    }
 
    // Driver function
    public static void main(String[] args)
    {
        System.out.println(convert0To5(10120));
    }
}
 
// This code is contributed by Kamal Rawal

Python

# Python program to replace all
# 0 with 5 in given integer
 
# A recursive function to replace all 0s
# with 5s in an integer
# Does'nt work if the given number is 0 itself
def convert0to5rec(num):
 
    # Base case for recurssion termination
    if num == 0:
        return 0
 
    # Extract the last digit and change it if needed
    digit = num % 10
 
    if digit == 0:
        digit = 5
 
    # Convert remaining digits and append the last digit
    return convert0to5rec(num / 10) * 10 + digit
 
# It handles 0 to 5 calls convert0to5rec()
# for other numbers
def convert0to5(num):
    if num == 0:
        return 5
    else:
        return convert0to5rec(num)
 
 
# Driver Program
num = 10120
print convert0to5(num)
 
# Contributed by Harshit Agrawal

C＃

// C# code for Replace all 0
// with 5 in an input Integer
using System;
 
class GFG {
 
    // A recursive function to replace
    // all 0s with 5s in an input number.
    // It doesn't work if input number
    // itself is 0.
    static int convert0To5Rec(int num)
    {
        // Base case
        if (num == 0)
            return 0;
 
        // Extraxt the last digit and
        // change it if needed
        int digit = num % 10;
        if (digit == 0)
            digit = 5;
 
        // Convert remaining digits
        // and append the last digit
        return convert0To5Rec(num / 10) * 10 + digit;
    }
 
    // It handles 0 and calls
    // convert0To5Rec() for other numbers
    static int convert0To5(int num)
    {
        if (num == 0)
            return 5;
        else
            return convert0To5Rec(num);
    }
 
    // Driver Code
    static public void Main()
    {
        Console.Write(convert0To5(10120));
    }
}
 
// This code is contributed by Raj

输出：

15125

复杂度分析：

• 时间复杂度： O(k)，重新函数仅调用k次，其中k是该数字的位数
• 空间复杂度： O(1)，不需要额外的空间。