我在理解分而治之算法时遇到了一些问题。我已经读过,为了成功地应用递归,你需要有一个“信仰的递归跳跃”,你不应该为每一步的细节而烦恼,但我真的不满意只接受递归有效条件得到满足,因为此刻对我来说似乎很神奇,我想知道它为什么会起作用。
所以我给出了以下在伪代码中寻找最大子阵列的递归算法:
Find-Maximum-Subarray(A, low, high)
if high == low
return (low, high, A[low])
else
mid = [(low + high)/2]
(left-low, left-high, left-sum) = Find-Maximum-Subarray(A, low, mid)
(right-low, right-high, right-sum) = Find-Maximum-Subarray(A,mid + 1, high)
(cross-low, cross-high, cross-sum) = Find-Max-Crossing-Subarray(A,low, mid, high)
if left-sum >= right-sum and left-sum >= cross-sum
return (left-low, left-high, left-sum)
else if right-sum >= left-sum and right-sum >= cross-sum
return (right-low, right-high, right-sum)
else
return (cross-low, cross-high, cross-sum)
其中Find-Max-Crossing-Subarray算法由以下伪代码给出:
Find-Maximum-Crossing-Subarray(A, low, mid, high)
left-sum = -INF
sum = 0
for i = mid down to low
sum = sum + A[i]
if sum > left-sum
left-sum = sum
max-left = i
right-sum = -INF
sum = 0
for j = mid + 1 to high
sum = sum + A[j]
if sum > right-sum
right-sum = sum
max-right = j
return (max-left, max-right, left-sum + right-sum)
现在,当我尝试将此算法应用于示例时,我很难理解所有步骤。
数组被“分解”(使用索引,而不实际更改数组本身)直到高等于低。我认为这对应于第一次调用,因此首先为数组左边的所有项调用Find-Maximum-Subarray,直到high == low == 1。然后返回(低,高,A [低]),在这种情况下将是(1,1,A [1])。现在我不明白在调用的其余部分中如何处理这些值。
此外,我不明白该算法实际上如何比较长度> 1的子阵列。任何人都可以向我解释一旦该函数的一个调用已经触底,该算法如何继续,请?
简而言之:
让A
成为n
长度的数组。你想要计算你称之为A
的Find-Maximum-Subarray(A, 0, n-1)
sou的最大子阵列。现在尝试使问题更容易:
A
切割成半长的数组B1
和B2
会发生什么。现在只有3个新病例
a)A
的Max子阵列也是B1
的子阵列但不是B2
的子阵列
b)A
的最大子阵列也是B2
的子阵列但不是B1
的子阵列
c)A
的最大子阵列与B1
和B2
重叠
所以你分别计算B1
和B2
的最大子阵列并寻找一个重叠的解决方案,最后你拿最大的一个。现在的诀窍是,你可以用B1
和B2
来做同样的事情。
例:
A =[-1, 2, -1, 1]
Call Find-Maximum-Subarray(A, 0, 3);
- Call Find-Maximum-Subarray(A, 0, 1); -> returns ( 1, 1, 2 ) (cause 2 > 1 > -1, see the subcalls)
- Call Find-Maximum-Subarray(A, 0, 0); -> returns ( 0, 0, -1 )
- Call Find-Maximum-Subarray(A, 1, 1); -> returns ( 1, 1, 2 )
- Call Find-Max-Crossing-Subarray(A, 0, 0, 1); -> returns ( 0, 1, 1 )
- Call Find-Maximum-Subarray(A, 2, 3); -> returns ( 3, 3, 1 ) ( 1 > 0 > -1, see subcalls)
- Call Find-Maximum-Subarray(A, 2, 2); -> returns ( 2, 2, -1 )
- Call Find-Maximum-Subarray(A, 3, 3); -> returns ( 3, 3, 1 )
- Call Find-Max-Crossing-Subarray(A, 2, 2, 3); returns ( 2, 3, 0 )
- Call Find-Max-Crossing-Subarray(A, 0, 1, 3); -> returns ( 1, 3, 2 )
- Here you have to take at least the elements A[1] and A[2] with the sum of 1,
but if you also take A[3]=1 the sum will be 2. taking A[0] does not help
due to A[0] is negative
- Now you have only to look which subarray has the larger sum. In this case you
have two with the same size: A[1] and A[1-3]. Return one of them.