拼多多笔试

def decrease(A, B):

    for i, v in enumerate(A):
        if i == 0:
            pre = True
            next = True if A[i + 1] - v > 0 else False
        elif i == len(A)-1:
            pre = True if v - A[i - 1] > 0 else False
            next = True
        else:
            pre = True if v - A[i-1]>0 else False
            next = True if A[i+1]-v>0 else False
        if pre and next == False:
            tal = i
            break
        else:
            tal = None

    if tal == None:
        return "NO"
    else:
        num=[]
        for j,v in enumerate(B):
            if tal == 0:
                if v<A[tal+1]:
                    num.append(v)
            elif tal == len(A)-1:
                if v>A[tal-1]:
                    num.append(v)
            else:
                if v>A[tal-1] and v<A[tal+1]:
                    num.append(v)
        A[tal] = max(num)
        return A
print(decrease([9,1,2,3,4,5,6],[1,2,0,5,-1]))

def solu(A):
    if len(A) == 0:
        return False

    i = 0
    j = 1
    while i<len(A):
        if i == len(A)-1:
            return A[i][-1] == A[0][0]
        else:
            if A[i][-1] == A[j][0]:
                if i+1 != j:
                    A[i+1],A[j] = A[j],A[i+1]
                i = i+1
                j = i+1
            else:
                j = j+1
                if j >= len(A):
                    return False

来源：https://blog.csdn.net/yl_mouse/article/details/98886793