在数组中进行二分查找的基本思想很简单,但是如果查找没有找到确切的项,它可能会返回一个“近似”索引(有时候我们可能会返回一个值大于或小于查找值的索引)。
为了寻找准确的插入点,似乎在我们得到近似位置之后,我们可能需要向左或向右“扫描”以寻找准确的插入位置,因此,比方说,在Ruby中,我们可以执行arr.insert(exact_index, value)
我有下面的解决方案,但是begin_index >= end_index
时的部分处理有点乱,不知道能不能用一个更优雅的解决方案?
(this如果找到了精确匹配,解决方案并不关心扫描多个匹配项,因此为精确匹配返回的索引可能指向与该值对应的任何索引......但我认为,如果它们都是整数,我们总是可以在知道找到了精确匹配项之后搜索a - 1
来找到左边界,或者搜索a + 1
来找到右边界。)
我的解决方案:
DEBUGGING = true
def binary_search_helper(arr, a, begin_index, end_index)
middle_index = (begin_index + end_index) / 2
puts "a = #{a}, arr[middle_index] = #{arr[middle_index]}, " +
"begin_index = #{begin_index}, end_index = #{end_index}, " +
"middle_index = #{middle_index}" if DEBUGGING
if arr[middle_index] == a
return middle_index
elsif begin_index >= end_index
index = [begin_index, end_index].min
return index if a < arr[index] && index >= 0 #careful because -1 means end of array
index = [begin_index, end_index].max
return index if a < arr[index] && index >= 0
return index + 1
elsif a > arr[middle_index]
return binary_search_helper(arr, a, middle_index + 1, end_index)
else
return binary_search_helper(arr, a, begin_index, middle_index - 1)
end
end
# for [1,3,5,7,9], searching for 6 will return index for 7 for insertion
# if exact match is found, then return that index
def binary_search(arr, a)
puts "\nSearching for #{a} in #{arr}" if DEBUGGING
return 0 if arr.empty?
result = binary_search_helper(arr, a, 0, arr.length - 1)
puts "the result is #{result}, the index for value #{arr[result].inspect}" if DEBUGGING
return result
end
arr = [1,3,5,7,9]
b = 6
arr.insert(binary_search(arr, b), b)
p arr
arr = [1,3,5,7,9,11]
b = 6
arr.insert(binary_search(arr, b), b)
p arr
arr = [1,3,5,7,9]
b = 60
arr.insert(binary_search(arr, b), b)
p arr
arr = [1,3,5,7,9,11]
b = 60
arr.insert(binary_search(arr, b), b)
p arr
arr = [1,3,5,7,9]
b = -60
arr.insert(binary_search(arr, b), b)
p arr
arr = [1,3,5,7,9,11]
b = -60
arr.insert(binary_search(arr, b), b)
p arr
arr = [1]
b = -60
arr.insert(binary_search(arr, b), b)
p arr
arr = [1]
b = 60
arr.insert(binary_search(arr, b), b)
p arr
arr = []
b = 60
arr.insert(binary_search(arr, b), b)
p arr
结果:
Searching for 6 in [1, 3, 5, 7, 9]
a = 6, arr[middle_index] = 5, begin_index = 0, end_index = 4, middle_index = 2
a = 6, arr[middle_index] = 7, begin_index = 3, end_index = 4, middle_index = 3
a = 6, arr[middle_index] = 5, begin_index = 3, end_index = 2, middle_index = 2
the result is 3, the index for value 7
[1, 3, 5, 6, 7, 9]
Searching for 6 in [1, 3, 5, 7, 9, 11]
a = 6, arr[middle_index] = 5, begin_index = 0, end_index = 5, middle_index = 2
a = 6, arr[middle_index] = 9, begin_index = 3, end_index = 5, middle_index = 4
a = 6, arr[middle_index] = 7, begin_index = 3, end_index = 3, middle_index = 3
the result is 3, the index for value 7
[1, 3, 5, 6, 7, 9, 11]
Searching for 60 in [1, 3, 5, 7, 9]
a = 60, arr[middle_index] = 5, begin_index = 0, end_index = 4, middle_index = 2
a = 60, arr[middle_index] = 7, begin_index = 3, end_index = 4, middle_index = 3
a = 60, arr[middle_index] = 9, begin_index = 4, end_index = 4, middle_index = 4
the result is 5, the index for value nil
[1, 3, 5, 7, 9, 60]
Searching for 60 in [1, 3, 5, 7, 9, 11]
a = 60, arr[middle_index] = 5, begin_index = 0, end_index = 5, middle_index = 2
a = 60, arr[middle_index] = 9, begin_index = 3, end_index = 5, middle_index = 4
a = 60, arr[middle_index] = 11, begin_index = 5, end_index = 5, middle_index = 5
the result is 6, the index for value nil
[1, 3, 5, 7, 9, 11, 60]
Searching for -60 in [1, 3, 5, 7, 9]
a = -60, arr[middle_index] = 5, begin_index = 0, end_index = 4, middle_index = 2
a = -60, arr[middle_index] = 1, begin_index = 0, end_index = 1, middle_index = 0
a = -60, arr[middle_index] = 9, begin_index = 0, end_index = -1, middle_index = -1
the result is 0, the index for value 1
[-60, 1, 3, 5, 7, 9]
Searching for -60 in [1, 3, 5, 7, 9, 11]
a = -60, arr[middle_index] = 5, begin_index = 0, end_index = 5, middle_index = 2
a = -60, arr[middle_index] = 1, begin_index = 0, end_index = 1, middle_index = 0
a = -60, arr[middle_index] = 11, begin_index = 0, end_index = -1, middle_index = -1
the result is 0, the index for value 1
[-60, 1, 3, 5, 7, 9, 11]
Searching for -60 in [1]
a = -60, arr[middle_index] = 1, begin_index = 0, end_index = 0, middle_index = 0
the result is 0, the index for value 1
[-60, 1]
Searching for 60 in [1]
a = 60, arr[middle_index] = 1, begin_index = 0, end_index = 0, middle_index = 0
the result is 1, the index for value nil
[1, 60]
Searching for 60 in []
[60]
3条答案
按热度按时间x759pob21#
这是Oracle Java中包含的Java的java.util.Arrays.binarySearch中的代码:
该算法已被证明是适当的,我喜欢这样的事实,即您可以立即从结果中知道它是一个精确匹配还是插入点的提示。
下面是我将其翻译成ruby的方法:
代码返回的值与OP为其测试数据提供的值相同。
mdfafbf12#
更新2020
其实二分法的插入问题已经有了很好的研究,有左插入点和右插入点,代码可以在Wikipedia和Rosetta Code上找到,比如要找到左插入点,代码是:
需要注意的是the overflow bug,所以
mid
确实应该被发现为low + floor((high - low) / 2)
。之前的答案:
实际上,使用
begin_index > end_index
可以更好地处理它,而不是检查begin_index >= end_index
,而且解决方案要干净得多:而且使用迭代代替递归可能更快,也更少担心堆栈溢出。
h7appiyu3#
左侧插入样品:
右侧插入样品:
样品存在:
样本测试用例:arr = [1、2、3、4、5、5、5、5、5、6、7、8、9、10]
结果=二进制搜索(arr,5)