修改二进制搜索[关闭]-Java 学习之路

-2

我一直试图解决一个问题，我有一个数组让我们说A [] = {5,9,11,15};和2个带有值的变量可以说2和10 .i需要找出数组的任何元素是否属于（2,10），即它的值在2（排除）到10（包括）之间 . 我可以简单地转一个循环搜索是否> if（2 = A [i]）<但这不会在数组大小的值上工作让我们说10 ^ 5.并且我尝试使用修改后的二进制搜索，它返回索引值小于或等于value的值密钥提供但失败 . 可以任何人为我提供快速算法吗？编辑：pos这里的元素数量breaaking是数组FLOOR是函数（修改二进制）

int Floor(int A[], int l, int r, int key)
{
int m;
while( r - l > 1 )
{
    m = l + (r - l)/2;

    if( A[m] <= key )
        l = m;
    else
        r = m;
}

return l;

}

int Floor（int A []，int size，int key）

{//错误检查

if（key <A [0]）返回-1; //没有底价

return Floor(A, 0, size, key);

}

int ret=Floor(breaaking,pos,mini);

 printf("%d\n",ret);

   printf("mini is %d and maxi is %d",mini,maxi);

    if(pos==0)
{  
    printf("There is no breaking point in the array :) (pos==0)\n");
    printf("Yes\n"); 
}
else if(ret==-1)
 { 
    printf("Mini is smaller than smallest element of breaking\n");

        if(breaaking[0]<maxi)
        { 

             printf("but maxi is greater than smallest element hece it lies between so:\n");
             printf("No\n");
        }
     else {
               printf("even maxi is less than smallest element hence:\n");
               printf("Yes\n");
           }
}
else if(ret==pos-1)
{ 
    printf("mini is either equal to last element of breaker set or greater than it\n");

       if(mini==breaaking[pos-1])
        {
             printf("mini is equal to the last element hence\n");    
                  printf("No\n");}
     else
        { 
        printf("mini is greater than the last element hence:"); 
         printf("Yes\n");
       }
 }
else
 { 
     printf("returned a valid index which is less than or equal to mini which is btw %d\n",ret);

        if(breaaking[ret]==mini)
        {
                 printf("mini was equal to one of the element of array hence\n"); 
                 printf("No\n");
        }
    else
    {  printf("mini is smaller than this element but greater than next element\n");
        if(breaaking[ret+1]<maxi)
             { 
                printf("next element lies between mini and maxi hence:\n")  ;      
                printf("No\n"); 
             }
        else
             {   printf("even maxi is smaller than next element hence\n");
                printf("Yes\n");
             }
     }
`}

1 回答

您只需使用 std::lower_bound 返回包含所有值的范围即可 . 如果没有，则范围将为空 .

#include <iostream>
#include <algorithm>
#include <tuple>

template<typename ForwardIterator>
std::pair<ForwardIterator, ForwardIterator>
range_inside(ForwardIterator b, ForwardIterator end, 
                  int lower, int upper) {
  auto it = std::lower_bound(b, end, lower);
  auto it2 = std::upper_bound(it, end, upper);
  return std::make_pair(it, it2);
}

int main()
{
  int arr[] = { 2, 5, 9, 10, 11, 15};
  int *r, *e;
  std::tie(r, e) = range_inside(std::begin(arr), std::end(arr), 2, 10);
  std::for_each(r, e, [] (int& x) { std::cout << x << " "; }); 
  // output: 2 5 9

  return 0;
}

回复于 2024-05-09T23:02:32+08:00

修改二进制搜索[关闭]

1 回答

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