我想修改着名的binary search算法,以返回下一个较大项目的索引而不是被搜索的密钥。
所以我们有4个案例:
e.g:
data = { 1, 3, 5, 7, 9, 11 };
搜索5或6个返回3。
while (low <= high) {
mid = (low + high) / 2;
if (data[mid] < val)
low = mid + 1;
else if (data[mid] > val)
high = mid - 1;
else {
break;
}
}
目前通过检查低值和高值来实现它。 有没有有趣的代码可以这样做!
编辑!!!
这是我如何运作:
if (low <= high)
found = (low + high) / 2 + 1;
else if (low >= data.length)
found = data.length ;
else if (high < 0)
found = -1;
else
found = low;
我正在寻找一种更优雅的方式!
编辑II !!!
如果没有重复,此代码可以正常工作。 为了处理重复的情况,我们需要修改第一个if条件:
if (low <= high)
found = (low + high) / 2 + 1;
迭代直到找到更大的元素。
答案 0 :(得分:5)
以下是满足OP搜索要求的一些C代码:
它还演示了4种不同类型的二进制搜索:
(假设data
)
#include <stdio.h>
int BinarySearch( int key, int data[], const int len )
{
int low = 0;
int high = len-1;
while( high >= low )
{
int mid = low + ((high - low) / 2);
/**/ if (data[mid] < key) low = mid + 1;
else if (data[mid] > key) high = mid - 1;
else return mid ;
}
return -1; // KEY_NOT_FOUND
}
int LessThanEqualBinSearch( int key, int data[], const int len )
{
int min = 0;
int max = len-1;
// var max = data.length - 1; // Javascript, Java conversion
while( min <= max)
{
int mid = min + ((max - min) / 2);
/**/ if (data[mid] < key) min = mid + 1;
else if (data[mid] > key) max = mid - 1;
else /*data[mid] = key)*/return mid ;
}
if( max < 0 )
return 0; // key < data[0]
else
if( min > (len-1))
return -1; // key >= data[len-1] // KEY_NOT_FOUND
else
return (min < max)
? min
: max + 1;
}
int LessThanEqualOrLastBinSearch( int key, int data[], const int len )
{
int min = 0;
int max = len-1;
// var max = data.length - 1; // Javascript, Java conversion
while( min <= max)
{
int mid = min + ((max - min) / 2);
/**/ if (data[mid] < key) min = mid + 1;
else if (data[mid] > key) max = mid - 1;
else /*data[mid] = key)*/return mid ;
}
if( max < 0 )
return 0; // key < data[0]
else
if( min > (len-1))
return len-1; // key >= data[len-1]
else
return (min < max)
? min
: max + 1;
}
int NextLargestBinSearch( int key, int data[], const int len )
{
int low = 0;
int high = len-1;
while( low <= high)
{
// To convert to Javascript:
// var mid = low + ((high - low) / 2) | 0;
int mid = low + ((high - low) / 2);
/**/ if (data[mid] < key) low = mid + 1;
else if (data[mid] > key) high = mid - 1;
else return mid + 1;
}
if( high < 0 )
return 0; // key < data[0]
else
if( low > (len-1))
return len; // key >= data[len-1]
else
return (low < high)
? low + 1
: high + 1;
}
int main()
{
int items[] = { 1, 3, 5, 7, 9, 11 };
int LENGTH = sizeof(items) / sizeof(items[0]);
for( int i = -1; i < 14; ++i )
printf( "[%2d]: == %2d <= %2d <| %d > %d\n", i
, BinarySearch ( i, items, LENGTH )
, LessThanEqualBinSearch ( i, items, LENGTH )
, LessThanEqualOrLastBinSearch( i, items, LENGTH )
, NextLargestBinSearch ( i, items, LENGTH )
);
return 0;
}
输出:
[-1]: == -1 <= 0 <| 0 > 0
[ 0]: == -1 <= 0 <| 0 > 0
[ 1]: == 0 <= 0 <| 0 > 1
[ 2]: == -1 <= 1 <| 1 > 1
[ 3]: == 1 <= 1 <| 1 > 2
[ 4]: == -1 <= 2 <| 2 > 2
[ 5]: == 2 <= 2 <| 2 > 3
[ 6]: == -1 <= 3 <| 3 > 3
[ 7]: == 3 <= 3 <| 3 > 4
[ 8]: == -1 <= 4 <| 4 > 4
[ 9]: == 4 <= 4 <| 4 > 5
[10]: == -1 <= 5 <| 5 > 5
[11]: == 5 <= 5 <| 5 > 6
[12]: == -1 <= -1 <| 5 > 6
[13]: == -1 <= -1 <| 5 > 6
1st
列是标准二进制搜索2nd
列是小于二进制搜索3rd
列是小于或最后二元搜索4th
列是下一个最大二进制搜索答案 1 :(得分:1)
这是一段代码:
public static int ceilSearch(int arr[], int low, int high, int x) {
int mid;
if (x <= arr[low])
return low;
if (x > arr[high])
return -1;
mid = (low + high) / 2; /* low + (high - low)/2 */
if (arr[mid] == x)
return mid;
else if (arr[mid] < x) {
if (mid + 1 <= high && x <= arr[mid + 1])
return mid + 1;
else
return ceilSearch(arr, mid + 1, high, x);
} else {
if (mid - 1 >= low && x > arr[mid - 1])
return mid;
else
return ceilSearch(arr, low, mid - 1, x);
}
}
答案 2 :(得分:0)
这是你想要的。它返回下一个更大的元素。
public int binarySearch(int[] arr, int key) {
int lo = 0;
int hi = arr.length - 1;int mid = 0;
while (lo <= hi) {
mid = (lo + hi) / 2;
if (key < arr[mid]) hi = mid - 1;
else if (key > arr[mid]) lo = mid + 1;
else return mid;
}
return -Math.min(lo, hi)-2;
}
public int myBinarySearch(int[] arr, int key){
int x = binarySearch(arr, key);
if(x >= arr.length-1 || -x > arr.length){
//whatever you want to return
return Integer.MAX_VALUE;
}
else if(x >= 0)
return arr[x+1] ;
else
return arr[-x-1];
}
public static void main(String args[]) {
Triall tr = new Triall();
int arr[] = { 1, 3, 5, 7, 9, 11 };
for( int i = 0; i < 13; i++ ) {
int n = tr.myBinarySearch( arr,i );
System.out.println(i + " " + n );
}
}