为什么二分搜索法优于三元搜索？
原文:https://www . geesforgeks . org/binary-search-preferred-三元-search/
下面是一个简单的 C++递归二分搜索法函数，取自这里的。
C
// A recursive binary search function. It returns location of x in
// given array arr[l..r] is present, otherwise -1
int binarySearch(int arr[], int l, int r, int x)
{
if (r >= l)
{
int mid = l + (r - l)/2;
// If the element is present at the middle itself
if (arr[mid] == x) return mid;
// If element is smaller than mid, then it can only be present
// in left subarray
if (arr[mid] > x) return binarySearch(arr, l, mid-1, x);
// Else the element can only be present in right subarray
return binarySearch(arr, mid+1, r, x);
}
// We reach here when element is not present in array
return -1;
}
Java 语言(一种计算机语言，尤用于创建网站)
// A recursive binary search function. It returns location of x in
// given array arr[l..r] is present, otherwise -1
static int binarySearch(int arr[], int l, int r, int x)
{
if (r >= l)
{
int mid = l + (r - l)/2;
// If the element is present at the middle itself
if (arr[mid] == x) return mid;
// If element is smaller than mid, then it can only be present
// in left subarray
if (arr[mid] > x) return binarySearch(arr, l, mid-1, x);
// Else the element can only be present in right subarray
return binarySearch(arr, mid+1, r, x);
}
// We reach here when element is not present in array
return -1;
}
// This code is contributed by gauravrajput1
Python 3
# A recursive binary search function. It returns location of x in
# given array arr[l..r] is present, otherwise -1
def binarySearch(arr, l, r, x):
if (r >= l):
mid = l + (r - l)/2;
# If the element is present at the middle itself
if (arr[mid] == x):
return mid;
# If element is smaller than mid, then it can only be present
# in left subarray
if (arr[mid] > x):
return binarySearch(arr, l, mid-1, x);
# Else the element can only be present in right subarray
return binarySearch(arr, mid+1, r, x);
# We reach here when element is not present in array
return -1;
# This code is contributed by umadevi9616
C
// A recursive binary search function. It returns location of x in
// given array arr[l..r] is present, otherwise -1
static int binarySearch(int []arr, int l, int r, int x)
{
if (r >= l)
{
int mid = l + (r - l)/2;
// If the element is present at the middle itself
if (arr[mid] == x) return mid;
// If element is smaller than mid, then it can only be present
// in left subarray
if (arr[mid] > x) return binarySearch(arr, l, mid-1, x);
// Else the element can only be present in right subarray
return binarySearch(arr, mid+1, r, x);
}
// We reach here when element is not present in array
return -1;
}
// This code is contributed by gauravrajput1
java 描述语言
<script>
// A recursive binary search function. It returns location of x in
// given array arr[l..r] is present, otherwise -1
function binarySearch(arr , l , r , x)
{
if (r >= l)
{
var mid = l + (r - l)/2;
// If the element is present at the middle itself
if (arr[mid] == x) return mid;
// If element is smaller than mid, then it can only be present
// in left subarray
if (arr[mid] > x) return binarySearch(arr, l, mid-1, x);
// Else the element can only be present in right subarray
return binarySearch(arr, mid+1, r, x);
}
// We reach here when element is not present in array
return -1;
}
// This code is contributed by gauravrajput1
</script>
下面是一个简单的递归三元搜索函数:
C
// A recursive ternary search function. It returns location of x in
// given array arr[l..r] is present, otherwise -1
int ternarySearch(int arr[], int l, int r, int x)
{
if (r >= l)
{
int mid1 = l + (r - l)/3;
int mid2 = mid1 + (r - l)/3;
// If x is present at the mid1
if (arr[mid1] == x) return mid1;
// If x is present at the mid2
if (arr[mid2] == x) return mid2;
// If x is present in left one-third
if (arr[mid1] > x) return ternarySearch(arr, l, mid1-1, x);
// If x is present in right one-third
if (arr[mid2] < x) return ternarySearch(arr, mid2+1, r, x);
// If x is present in middle one-third
return ternarySearch(arr, mid1+1, mid2-1, x);
}
// We reach here when element is not present in array
return -1;
}
Java 语言(一种计算机语言，尤用于创建网站)
import java.io.*;
class GFG
{
public static void main (String[] args)
{
// A recursive ternary search function.
// It returns location of x in given array
// arr[l..r] is present, otherwise -1
static int ternarySearch(int arr[], int l,
int r, int x)
{
if (r >= l)
{
int mid1 = l + (r - l) / 3;
int mid2 = mid1 + (r - l) / 3;
// If x is present at the mid1
if (arr[mid1] == x) return mid1;
// If x is present at the mid2
if (arr[mid2] == x) return mid2;
// If x is present in left one-third
if (arr[mid1] > x)
return ternarySearch(arr, l, mid1 - 1, x);
// If x is present in right one-third
if (arr[mid2] < x)
return ternarySearch(arr, mid2 + 1, r, x);
// If x is present in middle one-third
return ternarySearch(arr, mid1 + 1,
mid2 - 1, x);
}
// We reach here when element is
// not present in array
return -1;
}
}
Python 3
# A recursive ternary search function. It returns location of x in
# given array arr[l..r] is present, otherwise -1
def ternarySearch(arr, l, r, x):
if (r >= l):
mid1 = l + (r - l)//3
mid2 = mid1 + (r - l)//3
# If x is present at the mid1
if arr[mid1] == x:
return mid1
# If x is present at the mid2
if arr[mid2] == x:
return mid2
# If x is present in left one-third
if arr[mid1] > x:
return ternarySearch(arr, l, mid1-1, x)
# If x is present in right one-third
if arr[mid2] < x:
return ternarySearch(arr, mid2+1, r, x)
# If x is present in middle one-third
return ternarySearch(arr, mid1+1, mid2-1, x)
# We reach here when element is not present in array
return -1
# This code is contributed by ankush_953
C
// A recursive ternary search function.
// It returns location of x in given array
// arr[l..r] is present, otherwise -1
static int ternarySearch(int []arr, int l,
int r, int x)
{
if (r >= l)
{
int mid1 = l + (r - l) / 3;
int mid2 = mid1 + (r - l) / 3;
// If x is present at the mid1
if (arr[mid1] == x) return mid1;
// If x is present at the mid2
if (arr[mid2] == x) return mid2;
// If x is present in left one-third
if (arr[mid1] > x)
return ternarySearch(arr, l, mid1 - 1, x);
// If x is present in right one-third
if (arr[mid2] < x)
return ternarySearch(arr, mid2 + 1, r, x);
// If x is present in middle one-third
return ternarySearch(arr, mid1 + 1,
mid2 - 1, x);
}
// We reach here when element is
// not present in array
return -1;
}
// This code is contributed by gauravrajput1
服务器端编程语言（Professional Hypertext Preprocessor 的缩写）
<?php
// A recursive ternary search function.
// It returns location of x in
// given array arr[l..r] is
// present, otherwise -1
function ternarySearch($arr, $l,
$r, $x)
{
if ($r >= $l)
{
$mid1 = $l + ($r - $l) / 3;
$mid2 = $mid1 + ($r - l) / 3;
// If x is present at the mid1
if ($arr[mid1] == $x)
return $mid1;
// If x is present
// at the mid2
if ($arr[$mid2] == $x)
return $mid2;
// If x is present in
// left one-third
if ($arr[$mid1] > $x)
return ternarySearch($arr, $l,
$mid1 - 1, $x);
// If x is present in right one-third
if ($arr[$mid2] < $x)
return ternarySearch($arr, $mid2 + 1,
$r, $x);
// If x is present in
// middle one-third
return ternarySearch($arr, $mid1 + 1,
$mid2 - 1, $x);
}
// We reach here when element
// is not present in array
return -1;
}
// This code is contributed by anuj_67
?>
java 描述语言
<script>
// A recursive ternary search function.
// It returns location of x in given array
// arr[l..r] is present, otherwise -1
function ternarySearch(arr , l , r , x) {
if (r >= l) {
var mid1 = l + parseInt((r - l) / 3);
var mid2 = mid1 + parseInt((r - l) / 3);
// If x is present at the mid1
if (arr[mid1] == x)
return mid1;
// If x is present at the mid2
if (arr[mid2] == x)
return mid2;
// If x is present in left one-third
if (arr[mid1] > x)
return ternarySearch(arr, l, mid1 - 1, x);
// If x is present in right one-third
if (arr[mid2] < x)
return ternarySearch(arr, mid2 + 1, r, x);
// If x is present in middle one-third
return ternarySearch(arr, mid1 + 1, mid2 - 1, x);
}
// We reach here when element is
// not present in array
return -1;
// This code is contributed by gauravrajput1
</script>
在最坏的情况下，以上两者哪个做的比较少？ 从第一眼看上去，三元搜索进行比较的次数似乎较少，因为它进行 Log 3 n 次递归调用，但二分搜索法进行 Log 2 n 次递归调用。让我们仔细看看。 以下是二分搜索法最坏情况下计数比较的递归公式。
T(n) = T(n/2) + 2, T(1) = 1
以下是三值搜索最坏情况下计数比较的递归公式。
T(n) = T(n/3) + 4, T(1) = 1
在二分搜索法，最坏的情况下有 2 个对数 2 个 n + 1 的比较。在三元搜索中，最坏情况下有 4Log 3 n + 1 个比较。
Time Complexity for Binary search = 2clog2n + O(1)
Time Complexity for Ternary search = 4clog3n + O(1)
因此，三进制和二进制搜索的比较归结为表达式 2Log 3 n 和 Log 2 n 的比较。2Log 3 n 的值可以写成(2 / Log 2 3) * Log 2 n。由于(2 / Log 2 3)的值不止一个，所以在最坏的情况下，三进制搜索比二分搜索法做更多的比较。 练习: 为什么合并排序将输入数组分成两半，为什么不分成三个或更多部分？ 本文由安摩尔供稿。如果您发现任何不正确的地方，请写评论，或者您想分享更多关于上面讨论的主题的信息
版权属于：月萌API www.moonapi.com，转载请注明出处
