01-复杂度1 最大子列和问题_01-复杂度1 最大子列和问题分而治之-优快云博客

本文链接：https://blog.youkuaiyun.com/aoisbhc/article/details/102638927

本文介绍了三种求解最大子数组和的经典算法：暴力法、分而治之法和在线处理法。通过对比不同算法的时间复杂度，展示了如何从低效到高效地解决同一问题。文章详细解释了每种算法的工作原理，并提供了Python实现代码。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

#暴力
def m_s_c1(l):
    maxsum = 0
    start = 0
    end = 0
    for x in range(len(l)):
        thissum = 0
        for y in range(x,len(l)):
            thissum += l[y]
            if thissum > maxsum:
                maxsum = thissum
                start = x
                end = y
    return maxsum,start,end

#分而治之
def m_s_c2(list,left,right):
    #只有两个数
    if left == right:
        if list[left] > 0:
            return list[left]
        else:
            return 0

    #分
    mid = (left+right)//2

    #递归求左右两边子列最大和
    MaxLeftSum = m_s_c2(list,left,mid)
    MaxRightSum = m_s_c2(list,mid+1,right)

    #跨界最大子列和

    #从中线往左扫描
    MaxLeftBorderSum = 0
    LeftBorderSum = 0
    for index in range(mid,left-1,-1):  #注意遍历方向
        LeftBorderSum += list[index]
        if LeftBorderSum > MaxLeftBorderSum:
            MaxLeftBorderSum = LeftBorderSum

    # 从中线往右扫描
    MaxRightBorderSum = 0
    RightBorderSum = 0
    for index in range(mid+1,right+1):
        RightBorderSum += list[index]
        if RightBorderSum > MaxRightBorderSum:
            MaxRightBorderSum = RightBorderSum

    #治
    return max(MaxLeftSum,MaxRightSum,MaxLeftBorderSum+MaxRightBorderSum)

#在线处理
def m_s_c3(list):
    MaxSum = 0
    ThisSum = 0
    for index in range(len(list)):
        ThisSum += list[index]
        if ThisSum > MaxSum:
            MaxSum = ThisSum
        if ThisSum < 0:
            ThisSum = 0
    return MaxSum

def test1(l):
    # l = generate_list(n)
    res = m_s_c3(l)
    print(res)
K = input()
line = input()
l = list(map(lambda x:int(x),line.split()))
test1(l)