预期时间和给定时间之间的时间差

给定初始时钟时间h1:m1和当前时钟时间h2:m2 ，表示24 小时制格式的小时和分钟。当前时钟时间h2:m2可能正确也可能不正确。还给出了一个变量 K，它表示经过的小时数。任务是以秒为单位计算延迟，即预期时间和给定时间之间的时间差。
例子 ：

Input: h1 = 10, m1 = 12, h2 = 10, m2 = 17, k = 2 
Output: 115 minutes 
The clock initially displays 10:12. After 2 hours it must show 12:12. But at this point, the clock displays 10:17. Hence, the clock must be lagging by 115 minutes. so the answer is 115.
Input: h1 = 12, m1 = 00, h2 = 12, m2 = 58, k = 1 
Output: 2 minutes 
The clock initially displays 12:00. After 1 hour it must show 13:00. But at this point, the clock displays 12:58. Hence, the clock must be lagging by 2 minutes. so the answer is 2. 
 

方法：

1. 将 h:m 格式的给定时间转换为分钟数。它只是 60*h+m。
2. 计算两个计算时间（将 K 小时添加到初始时间）。
3. 找出分钟的差异，这将是答案。

下面是上述方法的实现。

C++

// C++ program to calculate clock delay
#include 
 
// Function definition with logic
int lagDuration(int h1, int m1, int h2, int m2, int k)
{
    int lag, t1, t2;
 
    // Conversion to minutes
    t1 = (h1 + k) * 60 + m1;
 
    // Conversion to minutes
    t2 = h2 * 60 + m2;
 
    // Calculating difference
    lag = t1 - t2;
    return lag;
}
 
// Driver Code
int main()
{
    int h1 = 12, m1 = 0;
    int h2 = 12, m2 = 58;
    int k = 1;
 
    int lag = lagDuration(h1, m1, h2, m2, k);
    printf("Lag = %d minutes", lag);
 
    return 0;
}

Java

// Java program to
// calculate clock delay
class GFG
{
 
// Function definition
// with logic
static int lagDuration(int h1, int m1,
                       int h2, int m2,
                       int k)
{
    int lag, t1, t2;
 
    // Conversion to minutes
    t1 = (h1 + k) * 60 + m1;
 
    // Conversion to minutes
    t2 = h2 * 60 + m2;
 
    // Calculating difference
    lag = t1 - t2;
    return lag;
}
 
// Driver Code
public static void main(String args[])
{
    int h1 = 12, m1 = 0;
    int h2 = 12, m2 = 58;
    int k = 1;
 
    int lag = lagDuration(h1, m1, h2, m2, k);
    System.out.println("Lag = " + lag +
                       " minutes");
}
}
 
// This code is contributed
// by Kirti_Mangal

Python3

# Python3 program to calculate clock delay
 
# Function definition with logic
def lagDuration(h1, m1, h2, m2, k):
    lag, t1, t2 = 0, 0, 0
     
    # Conversion to minutes
    t1 = (h1 + k) * 60 + m1
     
    # Conversion to minutes
    t2 = h2 * 60 + m2
     
    # Calculating difference
    lag = t1 - t2
    return lag
 
# Driver Code
h1, m1 = 12, 0
h2, m2 = 12, 58
k = 1
 
lag = lagDuration(h1, m1, h2, m2, k)
print("Lag =", lag, "minutes")
 
# This code has been contributed
# by 29AjayKumar

C#

// C# program to
// calculate clock delay
using System;
 
class GFG
{
 
// Function definition
// with logic
static int lagDuration(int h1, int m1,
                       int h2, int m2,
                       int k)
{
    int lag, t1, t2;
 
    // Conversion to minutes
    t1 = (h1 + k) * 60 + m1;
 
    // Conversion to minutes
    t2 = h2 * 60 + m2;
 
    // Calculating difference
    lag = t1 - t2;
    return lag;
}
 
// Driver Code
public static void Main()
{
    int h1 = 12, m1 = 0;
    int h2 = 12, m2 = 58;
    int k = 1;
 
    int lag = lagDuration(h1, m1, h2, m2, k);
    Console.WriteLine("Lag = " + lag +
                      " minutes");
}
}
 
// This code is contributed
// by Akanksha Rai(Abby_akku)

---

Javascript

// javascript program to
// calculate clock delay   
 
// Function definition
    // with logic
    function lagDuration(h1 , m1 , h2 , m2 , k) {
        var lag, t1, t2;
 
        // Conversion to minutes
        t1 = (h1 + k) * 60 + m1;
 
        // Conversion to minutes
        t2 = h2 * 60 + m2;
 
        // Calculating difference
        lag = t1 - t2;
        return lag;
    }
 
    // Driver Code
     
        var h1 = 12, m1 = 0;
        var h2 = 12, m2 = 58;
        var k = 1;
 
        var lag = lagDuration(h1, m1, h2, m2, k);
        document.write("Lag = " + lag + " minutes");
 
// This code is contributed by aashish1995

输出：

Lag = 2 minutes

时间复杂度：O(1)