arrays - 通过递归查找数组中的最大值

Question

// Find a maximum element in the array.
findMax(A)
   findMaxHelper(A, 0, A.length)

findMaxHelper(A, left, right)
   if (left == right - 1) 
      return A[left]
   else
      max1 = findMaxHelper(A, left, (right + left) / 2)
      max2 = findMaxHelper(A, (right + left) / 2, right)

      if (max1 > max2) 
         return max1 
      else 
         return max2

我很难理解这个伪代码中发生了什么。

有人可以帮助解释每行发生的事情。在回答问题之前，我需要理解这段代码。

我知道函数findMax调用辅助函数findMaxHelper，然后findMaxHelper使用递归。除此之外，我真的不明白。

Answer 1

You are using Divide and Conquer algorithm for finding the maximum element from the array. First, you are dividing the array into individual elements (divide), then you are comparing the elements (conquer). You are dividing the array using calling findMaxHelper recursively.

The general idea of Divide and conquer is shown in the figure:

Example:

Here max is same as your findMaxHelper function with two arguments i.e. left and right.

Check this example for more in depth understanding of the concept.

Answer 2

捷豹很好地提出了这个概念，保罗提供了正确而详细的解释。除此之外，我想分享一个简单的 C 代码，让您了解代码是如何执行的。这是 Jaguar 使用的相同输入的代码：

#include<stdio.h>
int findMaxHelper(int A[], int left, int right){
   int max1,max2;
   int static tabcount;
   int loop;
   for(loop = 0 ; loop <tabcount;loop++) printf("\t");
   tabcount++;
   printf(" Entering: findMaxHelper(A, left = %d ,right = %d)\n\n",left,right);
   if (left == right - 1){ 
      for(loop = 0 ; loop <tabcount;loop++) printf("\t");
      printf("\b\b\b\b\b\b\bLeaving: findMaxHelper(A, left = %d ,right = %d)| returning %d\n\n",left,right , A[left]);
      tabcount--;
      return A[left];
   }
   else
   {
      max1 = findMaxHelper(A, left, (right + left) / 2);
      max2 = findMaxHelper(A, (right + left) / 2, right);

      if (max1 > max2){ 
    for(loop = 0 ; loop <tabcount;loop++) printf("\t");
    printf("\b\b\b\b\b\b\bLeaving: findMaxHelper(A, left = %d ,right = %d) | returning max1=%d\n\n",left,right,max1);
    tabcount--;
    return max1;
    }
      else {
     for(loop = 0 ; loop <tabcount;loop++) printf("\t");
     printf("\b\b\b\b\b\b\bLeaving: findMaxHelper(A, left = %d ,right = %d)| returning max2=%d\n\n",left,right,max2);
     tabcount--;
     return max2;
    }

   }
}

int main (){
    int A[] = { 34,3,47,91,32,0 };
    int Ans =findMaxHelper(A,0,7);  
    printf( "And The Answer Is = %d \n",Ans);
}

您可以将代码复制粘贴到您的 linux 机器上...也许在每个 printf 之后放置 sleep(5) 并查看递归实际上是如何工作的！...希望这会有所帮助...我还将在这里分享我的系统的输出：

Entering: findMaxHelper(A, left = 0 ,right = 7)

     Entering: findMaxHelper(A, left = 0 ,right = 3)

         Entering: findMaxHelper(A, left = 0 ,right = 1)

         Leaving: findMaxHelper(A, left = 0 ,right = 1)| returning 34

         Entering: findMaxHelper(A, left = 1 ,right = 3)

             Entering: findMaxHelper(A, left = 1 ,right = 2)

             Leaving: findMaxHelper(A, left = 1 ,right = 2)| returning 3

             Entering: findMaxHelper(A, left = 2 ,right = 3)

             Leaving: findMaxHelper(A, left = 2 ,right = 3)| returning 47

         Leaving: findMaxHelper(A, left = 1 ,right = 3)| returning max2=47

     Leaving: findMaxHelper(A, left = 0 ,right = 3)| returning max2=47

     Entering: findMaxHelper(A, left = 3 ,right = 7)

         Entering: findMaxHelper(A, left = 3 ,right = 5)

             Entering: findMaxHelper(A, left = 3 ,right = 4)

             Leaving: findMaxHelper(A, left = 3 ,right = 4)| returning 91

             Entering: findMaxHelper(A, left = 4 ,right = 5)

             Leaving: findMaxHelper(A, left = 4 ,right = 5)| returning 32

         Leaving: findMaxHelper(A, left = 3 ,right = 5) | returning max1=91

         Entering: findMaxHelper(A, left = 5 ,right = 7)

             Entering: findMaxHelper(A, left = 5 ,right = 6)

             Leaving: findMaxHelper(A, left = 5 ,right = 6)| returning 0

             Entering: findMaxHelper(A, left = 6 ,right = 7)

             Leaving: findMaxHelper(A, left = 6 ,right = 7)| returning 0

         Leaving: findMaxHelper(A, left = 5 ,right = 7)| returning max2=0

     Leaving: findMaxHelper(A, left = 3 ,right = 7) | returning max1=91

 Leaving: findMaxHelper(A, left = 0 ,right = 7)| returning max2=91

And The Answer Is = 91 

Answer 3

findMaxHelper每次将数组分成两半，并在left，right中找到最大值：

例如你有数组A = [1, 3, 5, 8]，调用findMax(A)-> findMaxHelper(A, 0, A.length)：

     max1 | max2
     1 3  | 5 8

max1|max2 | max1|max2
1   |3    | 5   |8

Answer 4

#include<stdio.h>
#include<stdlib.h>

int high,*a,i=0,n,h;
int max(int *);

int main()
{

    printf("Size of array: ");
    scanf("%d",&n);

    a=(int *)malloc(n*sizeof(int));         //dynamic allocation
    for(i=0;i<n;i++)
    {
        scanf("%d",(a+i));
    }
        i=0;
    high=*a;
    h=max(a);
    printf("The highest element is %d\n",h);
}

int max(int *a)
{

    if(i<n)
    {   
        if(*(a+i)>high)
        {high=*(a+i);}
    i++;
    max(a);                     //recursive call
    }

    return high;
}

Answer 5

Basically finding max in array is not recommended by recursion as it is not required. Divide and conquer algorithms(recursive) are more time costly. But even though if you want to use it, you can use my below algorithm. Basically, it brings the largest element of array at first position and has almost linear running time.(This algo is just a recursive-illusion though!):

        int getRecursiveMax(int arr[], int size){
          if(size==1){
                      return arr[0];
          }else{
                 if(arr[0]< arr[size-1]){
                                      arr[0]=arr[size-1];
                     }
                 return(getRecursiveMax(arr,size-1));
            }

          }