鉴于城市的数n为n-1,其中站所在城市从0,任务是找到任何城市和它最近的车站之间的最大距离。请注意,带有车站的城市可以以任何顺序给出。
例子:
Input: numOfCities = 6, stations = [1, 4]
Output: 1
Input: numOfCities = 6, stations = [3]
Output: 3
Input: numOfCities = 6, stations = [3, 1]
Output: 2
- 下图代表第一个示例,其中包含6个城市以及车站以绿色突出显示的城市。在这种情况下,距离其最近车站最远的城市分别为0、2、3和5,距离为1。因此,最大距离为1。
- 在第二个示例中,离其最近车站最远的城市为0,距离为3。因此,最大距离为3。
- 在第三个示例中,距离其最近车站最远的城市为5,距离为2。因此,最大距离为2。
方法:此问题有三种可能的情况:
- 当最远的城市在两个车站之间。
- 当最远的城市在第一个车站的左侧。
- 当最远的城市在最后一个车站的右侧。
以下是解决上述问题的算法:
- 使用False初始化大小为n (城市数)的布尔数组。然后将带有桩号的城市的值标记为True
- 用0初始化变量dist 。初始化另一个变量maxDist ,其值等于带桩号的第一个城市(用于情况2)。
- 开始一个接一个地遍历所有城市。
- 如果当前城市有车站,则将( dist +1)// 2和maxDist的最大值分配给maxDist (用于情况1)。另外,将0分配给dist 。
- 否则,增加dist 。
- 最后,返回dist和maxDist的最大值(用于情况3)。
下面是上述方法的实现:
C++
// C++ program to calculate the maximum
// distance between any city
// and its nearest station
#include
using namespace std;
// Function to calculate the maximum
// distance between any city and its nearest station
int findMaxDistance(int numOfCities,int station[],int n)
{
// Initialize boolean list
bool hasStation[numOfCities + 1] = {false};
// Assign True to cities containing station
for (int city = 0; city < n; city++)
{
hasStation[station[city]] = true;
}
int dist = 0;
int maxDist = INT_MAX;
for(int i = 0; i < n; i++)
{
maxDist = min(station[i],maxDist);
}
for (int city = 0; city < numOfCities; city++)
{
if (hasStation[city] == true)
{
maxDist = max((dist + 1) / 2, maxDist);
dist = 0;
}
else
dist += 1;
}
return max(maxDist, dist);
}
//Driver code
int main()
{
int numOfCities = 6;
int station[] = {3, 1};
int n = sizeof(station)/sizeof(station[0]);
cout << "Max Distance:" << findMaxDistance(numOfCities,
station, n);
}
//This code is contributed by Mohit Kumar 29
Java
// Java program to calculate the maximum
// distance between any city
// and its nearest station
import java.util.*;
class GFG
{
// Function to calculate the maximum
// distance between any city and its nearest station
static int findMaxDistance(int numOfCities,
int station[],int n)
{
// Initialize boolean list
boolean hasStation[] = new boolean[numOfCities + 1];
// Assign True to cities containing station
for (int city = 0; city < n; city++)
{
hasStation[station[city]] = true;
}
int dist = 0;
int maxDist = Integer.MAX_VALUE;
for(int i = 0; i < n; i++)
{
maxDist = Math.min(station[i],maxDist);
}
for (int city = 0; city < numOfCities; city++)
{
if (hasStation[city] == true)
{
maxDist = Math.max((dist + 1) / 2, maxDist);
dist = 0;
}
else
dist += 1;
}
return Math.max(maxDist, dist);
}
//Driver code
public static void main(String args[])
{
int numOfCities = 6;
int station[] = {3, 1};
int n = station.length;
System.out.println("Max Distance:"+
findMaxDistance(numOfCities,station, n));
}
}
// This code is contributed by
// Surendra_Gnagwar
Python
# Python3 code to calculate the maximum
# distance between any city and its nearest station
# Function to calculate the maximum
# distance between any city and its nearest station
def findMaxDistance(numOfCities, station):
# Initialize boolean list
hasStation = [False] * numOfCities
# Assign True to cities containing station
for city in station:
hasStation[city] = True
dist, maxDist = 0, min(station)
for city in range(numOfCities):
if hasStation[city] == True:
maxDist = max((dist + 1) // 2, maxDist)
dist = 0
else:
dist += 1
return max(maxDist, dist)
numOfCities = 6
station = [3, 1]
print("Max Distance:", findMaxDistance(numOfCities, station))
C#
// C# program to calculate the maximum
// distance between any city
// and its nearest station
using System;
class GFG
{
// Function to calculate the maximum
// distance between any city and its nearest station
static int findMaxDistance(int numOfCities,
int []station,int n)
{
// Initialize boolean list
bool []hasStation = new bool[numOfCities + 1];
// Assign True to cities containing station
for (int city = 0; city < n; city++)
{
hasStation[station[city]] = true;
}
int dist = 0;
int maxDist = int.MaxValue;
for(int i = 0; i < n; i++)
{
maxDist = Math.Min(station[i],maxDist);
}
for (int city = 0; city < numOfCities; city++)
{
if (hasStation[city] == true)
{
maxDist = Math.Max((dist + 1) / 2, maxDist);
dist = 0;
}
else
dist += 1;
}
return Math.Max(maxDist, dist);
}
// Driver code
public static void Main(String []args)
{
int numOfCities = 6;
int []station = {3, 1};
int n = station.Length;
Console.WriteLine("Max Distance:"+
findMaxDistance(numOfCities,station, n));
}
}
// This code has been contributed by 29AjayKumar
PHP
Javascript
输出:
Max Distance: 2
时间复杂度: O(n)
空间复杂度: O(n)
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