使时间回文所需的最少分钟数

给定字符串str ，它以 24 小时格式存储时间为“HH: MM” 。任务是找到需要添加以使时间回文的最小分钟数。

例子：

Input: str = “05:39” 
Output: 11 
Explanation: It takes 11 minutes for minute value to become 50, 05:50 is a palindromic time

Examples: 
Input: str = “13:31” 
Output: 0 
Explanation: Since, 13:31 is already palindromic therefore, 0 minutes is required

方法：
这个想法是贪婪地增加分钟值，直到时间值变成回文。运行 while 循环以增加分钟值并同时检查小时值和分钟值是否形成回文。
在增加分钟和小时值时，请确保在分钟值为60且小时值为24时检查基本条件。
下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Function to get the required minutes
int get_palindrome_time(string str)
{
    int hh, mm;
 
    // Storing hour and minute value
    // in integral form
    hh
        = (str[0] - 48) * 10
        + (str[1] - 48);
    mm
        = (str[3] - 48) * 10
        + (str[4] - 48);
 
    int requiredTime = 0;
 
    // Keep iterating till first digit
    // hour becomes equal to second
    // digit of minute and second digit
    // of hour becomes equal to first
    // digit of minute
    while (hh % 10 != mm / 10
        || hh / 10 != mm % 10) {
 
        ++mm;
 
        // If mins is 60, increase hour, and
        // reinitilialized to 0
        if (mm == 60) {
            mm = 0;
            ++hh;
        }
 
        // If hours is 60, reinitialized to 0
        if (hh == 24)
            hh = 0;
        ++requiredTime;
    }
 
    // Return the required time
    return requiredTime;
}
 
// Driver Code
int main()
{
    // Given Time as a string
    string str = "05:39";
 
    // Function Call
    cout << get_palindrome_time(str)
        << endl;
}

Java

// Java program for the above approach
class GFG{
     
// Function to get the required minutes
public static int get_palindrome_time(String str)
{
    int hh, mm;
 
    // Storing hour and minute value
    // in integral form
    hh = (str.charAt(0) - 48) * 10 +
        (str.charAt(1) - 48);
    mm = (str.charAt(3) - 48) * 10 +
        (str.charAt(4) - 48);
 
    int requiredTime = 0;
 
    // Keep iterating till first digit
    // hour becomes equal to second
    // digit of minute and second digit
    // of hour becomes equal to first
    // digit of minute
    while (hh % 10 != mm / 10 ||
        hh / 10 != mm % 10)
    {
        ++mm;
 
        // If mins is 60, increase hour, and
        // reinitilialized to 0
        if (mm == 60)
        {
            mm = 0;
            ++hh;
        }
 
        // If hours is 60, reinitialized to 0
        if (hh == 24)
            hh = 0;
        ++requiredTime;
    }
 
    // Return the required time
    return requiredTime;
}
 
// Driver code
public static void main(String[] args)
{
     
    // Given Time as a string
    String str = "05:39";
 
    // Function Call
    System.out.println(get_palindrome_time(str));
}
}
 
// This code is contributed by divyeshrabadiya07

Python3

# Python3 program for the above approach
 
# Function to get the required minutes 
def get_palindrome_time(str):
     
    # Storing hour and minute value
    # in integral form
    hh = ((ord(str[0]) - 48) * 10 +
          (ord(str[1]) - 48))
    mm = ((ord(str[3]) - 48) * 10 +
          (ord(str[4]) - 48))
     
    requiredTime = 0
 
    # Keep iterating till first digit
    # hour becomes equal to second
    # digit of minute and second digit
    # of hour becomes equal to first
    # digit of minute
    while (hh % 10 != mm // 10 or
          hh // 10 != mm % 10):
        mm += 1
 
        # If mins is 60, increase hour, and
        # reinitilialized to 0
        if (mm == 60):
            mm = 0
            hh += 1
 
        # If hours is 60, reinitialized to 0
        if (hh == 24):
            hh = 0
             
        requiredTime += 1;
         
    # Return the required time
    return requiredTime
     
if __name__=="__main__":
     
    # Given Time as a string
    str = "05:39";
 
    # Function call
    print(get_palindrome_time(str));
 
# This code is contributed by rutvik_56

C#

// C# program for the above approach
using System;
 
class GFG{
     
// Function to get the required minutes
public static int get_palindrome_time(string str)
{
    int hh, mm;
 
    // Storing hour and minute value
    // in integral form
    hh = (str[0] - 48) * 10 +
        (str[1] - 48);
    mm = (str[3] - 48) * 10 +
        (str[4] - 48);
 
    int requiredTime = 0;
 
    // Keep iterating till first digit
    // hour becomes equal to second
    // digit of minute and second digit
    // of hour becomes equal to first
    // digit of minute
    while (hh % 10 != mm / 10 ||
        hh / 10 != mm % 10)
    {
        ++mm;
 
        // If mins is 60, increase hour,
        // and reinitilialized to 0
        if (mm == 60)
        {
            mm = 0;
            ++hh;
        }
 
        // If hours is 60, reinitialized to 0
        if (hh == 24)
            hh = 0;
        ++requiredTime;
    }
 
    // Return the required time
    return requiredTime;
}
 
// Driver code
public static void Main(string[] args)
{
     
    // Given Time as a string
    string str = "05:39";
 
    // Function Call
    Console.Write(get_palindrome_time(str));
}
}
 
// This code is contributed by rutvik_56

Javascript

 
// Javascript program for the above approach
  
// Function to get the required minutes
function get_palindrome_time(str)
{
    let hh, mm;
  
    // Storing hour and minute value
    // in letegral form
    hh = (str[0].charCodeAt() - 48) * 10 +
        (str[1].charCodeAt() - 48);
    mm = (str[3].charCodeAt() - 48) * 10 +
        (str[4].charCodeAt() - 48);
  
    let requiredTime = 0;
  
    // Keep iterating till first digit
    // hour becomes equal to second
    // digit of minute and second digit
    // of hour becomes equal to first
    // digit of minute
    while (hh % 10 != Math.floor(mm / 10) ||
        Math.floor(hh / 10) != mm % 10)
    {
        ++mm;
  
        // If mins is 60, increase hour, and
        // reinitilialized to 0
        if (mm == 60)
        {
            mm = 0;
            ++hh;
        }
  
        // If hours is 60, reinitialized to 0
        if (hh == 24)
            hh = 0;
        ++requiredTime;
    }
  
    // Return the required time
    return requiredTime;
}
// Driver Code
 
    // Given Time as a string
    let str = "05:39";
  
    // Function Call
    document.write(get_palindrome_time(str.split('')));
 

输出：

11

时间复杂度： O(1)
辅助空间： O(1)

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