最大子数组问题——python

典型分治思想，连续子数组必然位于以下三种情况之中：

        1、完全位于A[low..mid]中，因此low <= i <= j <= mid.

        2、完全位于A[mid..high]中，因此mid <= i <= j <= high.

        3、跨越了中点，因此low <= i <= mid < j <= high.

求sum：

def max_child2(alist, low, high):
    if low == high:
        return alist[low]
    mid = (low + high) // 2
    m1 = max_child2(alist, low, mid)
    m2 = max_child2(alist, mid+1, high)
    
    now_left = alist[mid]
    left = alist[mid]
    for i in range(mid-1, low-1, -1):
        now_left = now_left + alist[i]
        if now_left > left:
            left = now_left
    now_right = alist[mid+1]
    right = alist[mid+1]
    for j in range(mid+2, high+1):
        now_right = now_right + alist[j]
        if now_right > right:
            right = now_right
    m3 = left + right
    result = max(m1,m2,m3)
    return result

求sum和index：

def MaxCrossSubArray(A,low,mid,high):
    LeftMaxSum=A[mid]
    leftSum=A[mid]
    leftIndex=mid
    for i in range(mid-1,low-1,-1):
        leftSum=leftSum+A[i]
        if leftSum>LeftMaxSum:
            LeftMaxSum=leftSum
            leftIndex=i
    rightMaxSum=0
    rightSum=0
    rightIndex=mid
    for i in range(mid+1,high+1):
        rightSum+=A[i]
        if rightSum>rightMaxSum:
            rightMaxSum=rightSum
            rightIndex=i
    MaxSum=LeftMaxSum+rightMaxSum
    return (MaxSum,leftIndex,rightIndex)

def MaxSubArray(A,low,high):
    if low==high:
        return (A[low],low,high)
    mid=(low+high)//2
    Left=MaxSubArray(A,low,mid)
    Cross=MaxCrossSubArray(A,low,mid,high)
    Right=MaxSubArray(A,mid+1,high)
    List=[Left,Cross,Right]
    result=sorted(List,key = lambda list : list[0],reverse=True)
    return result[0]

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