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📜  荷兰国家国旗

📅  最后修改于: 2020-12-10 06:37:06             🧑  作者: Mango

荷兰国家国旗

荷兰国旗(DNF)-这是Edsger Dijkstra提出的编程问题。荷兰的国旗由三种颜色组成：白色，红色和蓝色。任务是以相同颜色的球放置在一起的方式随机排列白色，红色和蓝色的球。对于DNF(荷兰国旗)，我们在线性时间内对数组0、1和2进行了排序，而不会占用任何额外的空间。我们必须记住，该算法只能在具有三个唯一元素的数组上实现。

算法-

  • 取三分，即-低，中，高。
  • 我们在开头使用低指针和中指针，高指针将指向给定数组的末尾。

案例：

  • 如果数组[mid] = 0，则将数组[mid]与数组[low]交换，并将两个指针递增一次。
  • 如果数组[mid] = 1，则不需要交换。一次增加中间指针。
  • 如果数组[mid] = 2，则我们将数组[mid]与数组[high]交换并减少一次高指针。

代码-

(C语言)

#include  
using namespace std;  
// Function to sort the input array where the array is assumed to have values in {0, 1, 2}
// We have to take 3 distint or unique elements 
void JTP(int arr[], int arr_size)  
{ 
int low = 0;  
int high = arr_size - 1;  
int mid = 0;  
// We have keep iterating till all the elements are sorted  
while (mid <= high)  
{  
switch (arr[mid])  
{  
// Here mid is 0.  
case 0:  
swap(arr[low++], arr[mid++]);  
break;  
// Here mid is 1.  
case 1:  
mid++;  
break;  
// Here mid is 2.  
case 2:  
swap(arr[mid], arr[high--]);  
break;  
}  
}  
}  
// Now, we write the function to print array arr[]  
void printArray(int arr[], int arr_size)  
{  
// To iterate and print every element, we follow these steps  
for (int i = 0; i < arr_size; i++)  
cout << arr[i] << " ";  
}  
//Main Code  
int main()  
{  
int arr[] = {0,1,0,1,2,0,1,2};  
int n = sizeof(arr)/sizeof(arr[0]);  
cout << "Array before executing the algorithm: ";  
printArray(arr, n);   
JTP(arr, n);  
cout << "\nArray after executing the DNFS algorithm: ";  
printArray(arr, n);  
return 0;  
}

输出-

执行算法之前的数组：0 1 0 1 2 0 1 2

执行DNFS算法后的数组：0 0 0 1 1 1 2 2

代码-

(在爪哇)

import java.io.*; 
class DNF { 
static void JTP(int arr[], int arr_size) 
{ 
int low = 0; 
int high = arr_size - 1; 
int mid = 0, temp=0; // We use temporary variable for swapping 
while (mid <= high) 
{ 
switch (arr[mid]) 
{
case 0: // Here mid pointer points is at 0.
{ 
temp = arr[low]; 
arr[low] = arr[mid]; 
arr[mid] = temp; 
low++; 
mid++; 
break;
} 
case 1: // Here mid pointer points is at 1.
mid++; 
break; 
case 2: // Here mid pointer points is at 2.
{ 
temp = arr[mid]; 
arr[mid] = arr[high]; 
arr[high] = temp; 
high--; 
break; 
} 
} 
} 
} 
// Now we have to call function to print array arr[] 
static void printArray(int arr[], int arr_size) 
{ 
int i; 
for (i = 0; i < arr_size; i++) 
System.out.print(arr[i]+" "); 
System.out.println(""); 
} 
//Now we use driver function to check for above functions
public static void main (String[] arguments) 
{ 
int arr[] = {0, 1, 0, 1, 2, 0, 1, 2}; 
int arr_size = arr.length; 
System.out.println("Array before executing the DNFS algorithm : "); 
printArray(arr, arr_size); 
JTP(arr, arr_size); 
System.out.println("\nArray after executing the DNFS algorithm : "); 
printArray(arr, arr_size); 
} 
}

输出-

执行DNFS算法之前的数组：0 1 0 1 2 0 1 2

执行DNFS算法后的数组：0 0 0 1 1 1 2 2 2

代码-

(在PYTHON中)

def JTP( arr, arr_size): 
  low = 0
  high = arr_size - 1
  mid = 0
  while mid <= high: 
    if arr[mid] == 0: 
      arr[low],arr[mid] = arr[mid],arr[low] 
      low = low + 1
      mid = mid + 1
    elif arr[mid] == 1: 
      mid = mid + 1
    else: 
      arr[mid],arr[high] = arr[high],arr[mid] 
      high = high - 1
  return arr 
    
# Function to print array 
def printArray(arr): 
    for k in arr: 
        print k, 
    print
    
# Driver Program 
arr = [0, 1, 0, 1, 1, 2, 0, 1, 2] 
arr_size = len(arr)
print " Array before executing the algorithm: ", 
printArray(arr) 
arr = JTP(arr, arr_size) 
print "Array after executing the DNFS algorithm: ", 
printArray(arr)

输出-

执行算法之前的数组：0 1 0 1 1 2 0 1 2

执行DNFS算法后的数组：0 0 0 1 1 1 1 2 2 2