算法导论-----分治策略----------求最大子数组

问题描述：

在一个包含负数的数组中，找出和最大的子数组。

算法描述：

使用分治策略，将数组划分为两个规模尽量相等的子数组。也就是找到数组的中央位置mid。然后考虑求解A[low..mid],A[mid+1,high]。

最大子数组必然为下列三种情况之一:

1.位于A[low..mid]，完全位于左数组

2.位于A[mid+1..high]，完全位于右数组

3.位于A[i..mid..j]，即跨越了中点，且 low<=i<=mid<=j<=high

python实现：

#跨越了中点的最大数组

def max_subArray_inMid(ary,low,mid,high):
max_sum = -100000
left_index=right_index = mid
sum = 0

i = mid
while low<=i:
sum+=ary[i]
if sum > max_sum:
max_sum = sum
left_index = i
i-=1

i = mid + 1
sum = max_sum
while i <= high:
sum+=ary[i]
if sum > max_sum:
max_sum = sum
right_index = i
i+=1

return (max_sum,left_index,right_index)

#没有跨越中点的最大数组
def max_subArray(ary,low,high):
if(low == high):
return (ary[low],low,high)

mid = int((low+high)/2)
(left_max,left_left,left_right) = max_subArray(ary,low,mid)
(right_max,right_left,right_right) = max_subArray(ary,mid+1,high)

(mid_max,mid_left,mid_right) = max_subArray_inMid(ary,low,mid,high)
if left_max > right_max and left_max > mid_max:
return (left_max,left_left,left_right)
elif right_max > left_max and right_max > mid_max:
return (right_max,right_left,right_right)
else:
return(mid_max,mid_left,mid_right)

ary = [13,-3,-25,20,-3,-16,-23,18,20,-7,12,-5,-22,15,-4,7]
print(max_subArray(ary,0,len(ary)-1))