计算与给定会议时间相交的时间间隔
原文:https://www . geesforgeks . org/count-intervals-与给定会议时间相交的时间间隔/
给定一个由代表开始和结束时间(以 12 小时格式)的NT6】对串和代表会议时间的串 P 组成的数组arr[]【2】，任务是找出包含时间的间隔计数P
示例:
输入:P =“12:01:AM”，arr[][2]= {“12:00:AM”、“11:55:PM”}、{“12:01:AM”、“11:50:AM”}、{“12:30:AM”、“12:00:PM”}、{“11:57:AM”、“11:59:PM”} } 输出: 2 解释:【T7
输入:P =“12:01:AM”，arr[][2]= { {“09:57:AM”，“12:00:PM”} T3】输出:0 T6】说明:无区间包含时间 P
进场:思路是先把所有时间从 12 小时格式转换成 24 小时格式然后和时间 P 进行范围对比。按照以下步骤解决问题:
- 首先将 12 小时格式的所有时间转换为 24 小时格式，然后存储 24 小时时间格式的整数值。
- 初始化一个变量，比如说 ans，来存储 P 所在的时间间隔的计数。
- 若 (P ≥ L & & P ≤ R) 或 (P ≥ R & & P ≤ L) ，则遍历数组arr【】并将 ans 的计数增加 1 。
- 最后，完成以上步骤后，打印 ans 。
下面是上述方法的实现:
C++
// C++ implementation of the above approach
#include <bits/stdc++.h>
using namespace std;
// Function to convert a time in 24
// hour format to an equivalent integer
int convert(string str)
{
// Removes ":" at 3rd position
str.replace(2, 1, "");
// Calculate hours
int h1 = (int)str[1] - '0';
int h2 = (int)str[0] - '0';
int hh = (h2 * 10 + h1 % 10);
// Stores the time in 24 hours format
int time = 0;
// If time is in "AM"
if (str[5] == 'A') {
// If hh is equal to 12
if (hh == 12)
time += stoi(str.substr(2, 2));
else {
time += stoi(str.substr(0, 2));
}
}
// If time is in "PM"
else {
// If hh is equal to 12
if (hh == 12) {
time += stoi(str.substr(0, 4));
}
else {
time += stoi(str.substr(0, 4));
time += 1200;
}
}
// Return time
return time;
}
// Function to count number
// of intervals in which p lies
int countOverlap(string arr[][2],
int n, string p)
{
// Stores the count
int ans = 0;
// Stores the integer value of
// 24 hours time format of P
int M = convert(p);
// Traverse the array
for (int i = 0; i < n; i++) {
// Stores the integer value of
// 24 hours time format of arr[i][0]
int L = convert(arr[i][0]);
// Stores the integer value of
// 24 hours time format of arr[i][1]
int R = convert(arr[i][1]);
// If M lies within the [L, R]
if ((L <= M && M <= R)
|| (M >= R && M <= L))
// Increment ans by 1
ans++;
}
// Return ans
return ans;
}
// Driver Code
int main()
{
string arr[][2] = { { "12:00:AM", "11:55:PM" },
{ "12:01:AM", "11:50:AM" },
{ "12:30:AM", "12:00:PM" },
{ "11:57:AM", "11:59:PM" } };
string P = "12:01:PM";
int N = sizeof(arr) / sizeof(arr[0]);
cout << countOverlap(arr, N, P) << endl;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java implementation of the above approach
import java.io.*;
class GFG
{
// Function to convert a time in 24
// hour format to an equivalent integer
static int convert(String str)
{
// Removes ":" at 3rd position
str = str.substring(0, 2) + str.substring(3);
// Calculate hours
int h1 = (int)str.charAt(1) - '0';
int h2 = (int)str.charAt(0) - '0';
int hh = (h2 * 10 + h1 % 10);
// Stores the time in 24 hours format
int time = 0;
// If time is in "AM"
if (str.charAt(5) == 'A') {
// If hh is equal to 12
if (hh == 12)
time += Integer.parseInt(
str.substring(2, 4));
else {
time += Integer.parseInt(
str.substring(0, 2));
}
}
// If time is in "PM"
else {
// If hh is equal to 12
if (hh == 12) {
time += Integer.parseInt(
str.substring(0, 4));
}
else {
time += Integer.parseInt(
str.substring(0, 4));
time += 1200;
}
}
// Return time
return time;
}
// Function to count number
// of intervals in which p lies
static int countOverlap(String arr[][], int n, String p)
{
// Stores the count
int ans = 0;
// Stores the integer value of
// 24 hours time format of P
int M = convert(p);
// Traverse the array
for (int i = 0; i < n; i++)
{
// Stores the integer value of
// 24 hours time format of arr[i][0]
int L = convert(arr[i][0]);
// Stores the integer value of
// 24 hours time format of arr[i][1]
int R = convert(arr[i][1]);
// If M lies within the [L, R]
if ((L <= M && M <= R) || (M >= R && M <= L))
// Increment ans by 1
ans++;
}
// Return ans
return ans;
}
// Driver Code
public static void main(String[] args)
{
String[][] arr
= new String[][] { { "12:00:AM", "11:55:PM" },
{ "12:01:AM", "11:50:AM" },
{ "12:30:AM", "12:00:PM" },
{ "11:57:AM", "11:59:PM" } };
String P = "12:01:PM";
int N = arr.length;
System.out.println(countOverlap(arr, N, P));
}
}
// This code is contributed by Dharanendra L V
C
// C# implementation of the above approach
using System;
class GFG{
// Function to convert a time in 24
// hour format to an equivalent integer
static int convert(String str)
{
// Removes ":" at 3rd position
str = str.Substring(0, 2) + str.Substring(3);
// Calculate hours
int h1 = (int)str[1] - '0';
int h2 = (int)str[0] - '0';
int hh = (h2 * 10 + h1 % 10);
// Stores the time in 24 hours format
int time = 0;
// If time is in "AM"
if (str[5] == 'A')
{
// If hh is equal to 12
if (hh == 12)
time += Int32.Parse(
str.Substring(2, 2));
else
{
time += Int32.Parse(
str.Substring(0, 2));
}
}
// If time is in "PM"
else
{
// If hh is equal to 12
if (hh == 12)
{
time += Int32.Parse(
str.Substring(0, 4));
}
else
{
time += Int32.Parse(
str.Substring(0, 4));
time += 1200;
}
}
// Return time
return time;
}
// Function to count number
// of intervals in which p lies
static int countOverlap(String [,]arr, int n,
String p)
{
// Stores the count
int ans = 0;
// Stores the integer value of
// 24 hours time format of P
int M = convert(p);
// Traverse the array
for(int i = 0; i < n; i++)
{
// Stores the integer value of
// 24 hours time format of arr[i,0]
int L = convert(arr[i,0]);
// Stores the integer value of
// 24 hours time format of arr[i,1]
int R = convert(arr[i,1]);
// If M lies within the [L, R]
if ((L <= M && M <= R) ||
(M >= R && M <= L))
// Increment ans by 1
ans++;
}
// Return ans
return ans;
}
// Driver Code
public static void Main(String[] args)
{
String[,] arr = new String[,]{
{ "12:00:AM", "11:55:PM" },
{ "12:01:AM", "11:50:AM" },
{ "12:30:AM", "12:00:PM" },
{ "11:57:AM", "11:59:PM" } };
String P = "12:01:PM";
int N = arr.GetLength(0);
Console.WriteLine(countOverlap(arr, N, P));
}
}
// This code is contributed by 29AjayKumar
Output:
2
时间复杂度:O(N) T5辅助空间:** O(1)
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