找到容纳乘客所需的列车最小载客量
给定进出列车的乘客数量，任务是找到列车的最小容量，以在整个旅程中保持所有乘客。 例:
输入:进入[] = {3，5，2，0}，退出[] = {0，2，4，4} 输出: 6 车站 1:列车通过能力= 3 车站 2:列车通过能力= 3+5–2 = 6 车站 3:列车通过能力= 6+2–4 = 4 车站 4:列车通过能力= 4–4 = 0 任何时刻 列车最多可容纳 6 名乘客。 输入:输入[] = {5，2，2，0}，退出[] = {0，2，2，5} 输出: 5
进场:某站列车当前通过能力可通过将列车进站人数相加，再减去出站人数计算得出。所需的最小容量将是所有站点当前容量的最大值。 以下是上述办法的实施情况:
C++
// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
// Function to return the minimum capacity required
int minCapacity(int enter[], int exit[], int n)
{
// To store the minimum capacity
int minCap = 0;
// To store the current capacity
// of the train
int currCap = 0;
// For every station
for (int i = 0; i < n; i++) {
// Add the number of people entering the
// train and subtract the number of people
// exiting the train to get the
// current capacity of the train
currCap = currCap + enter[i] - exit[i];
// Update the minimum capacity
minCap = max(minCap, currCap);
}
return minCap;
}
// Driver code
int main()
{
int enter[] = { 3, 5, 2, 0 };
int exit[] = { 0, 2, 4, 4 };
int n = sizeof(enter) / sizeof(enter[0]);
cout << minCapacity(enter, exit, n);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java implementation of the approach
import java.util.*;
class GFG
{
// Function to return the minimum capacity required
static int minCapacity(int enter[],
int exit[], int n)
{
// To store the minimum capacity
int minCap = 0;
// To store the current capacity
// of the train
int currCap = 0;
// For every station
for (int i = 0; i < n; i++)
{
// Add the number of people entering the
// train and subtract the number of people
// exiting the train to get the
// current capacity of the train
currCap = currCap + enter[i] - exit[i];
// Update the minimum capacity
minCap = Math.max(minCap, currCap);
}
return minCap;
}
// Driver code
public static void main(String[] args)
{
int enter[] = { 3, 5, 2, 0 };
int exit[] = { 0, 2, 4, 4 };
int n = enter.length;
System.out.println(minCapacity(enter, exit, n));
}
}
// This code is contributed by Rajput-Ji
Python 3
# Python3 implementation of the approach
# Function to return the
# minimum capacity required
def minCapacity(enter, exit, n):
# To store the minimum capacity
minCap = 0;
# To store the current capacity
# of the train
currCap = 0;
# For every station
for i in range(n):
# Add the number of people entering the
# train and subtract the number of people
# exiting the train to get the
# current capacity of the train
currCap = currCap + enter[i] - exit[i];
# Update the minimum capacity
minCap = max(minCap, currCap);
return minCap;
# Driver code
if __name__ == '__main__':
enter = [3, 5, 2, 0];
exit = [0, 2, 4, 4];
n = len(enter);
print(minCapacity(enter, exit, n));
# This code is contributed by Princi Singh
C
// C# implementation of the approach
using System;
class GFG
{
// Function to return the minimum
// capacity required
static int minCapacity(int []enter,
int []exit, int n)
{
// To store the minimum capacity
int minCap = 0;
// To store the current capacity
// of the train
int currCap = 0;
// For every station
for (int i = 0; i < n; i++)
{
// Add the number of people entering the
// train and subtract the number of people
// exiting the train to get the
// current capacity of the train
currCap = currCap + enter[i] - exit[i];
// Update the minimum capacity
minCap = Math.Max(minCap, currCap);
}
return minCap;
}
// Driver code
public static void Main(String[] args)
{
int []enter = { 3, 5, 2, 0 };
int []exit = { 0, 2, 4, 4 };
int n = enter.Length;
Console.WriteLine(minCapacity(enter, exit, n));
}
}
// This code is contributed by PrinciRaj1992
java 描述语言
<script>
// Javascript implementation of the approach
// Function to return the minimum capacity required
function minCapacity(enter, exit, n) {
// To store the minimum capacity
let minCap = 0;
// To store the current capacity
// of the train
let currCap = 0;
// For every station
for (let i = 0; i < n; i++) {
// Add the number of people entering the
// train and subtract the number of people
// exiting the train to get the
// current capacity of the train
currCap = currCap + enter[i] - exit[i];
// Update the minimum capacity
minCap = Math.max(minCap, currCap);
}
return minCap;
}
// Driver code
let enter = [3, 5, 2, 0];
let exit = [0, 2, 4, 4];
let n = enter.length;
document.write(minCapacity(enter, exit, n));
// This code is contributed by _saurabh_jaiswal.
</script>
Output:
6
时间复杂度: O(n)
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