乐鑫笔试编程题

题意：存在一根电线杆，可以是第一根或者最后一根，在该电线杆前的电线杆的高度是不增的，在它之后的电线杆的高度是不减的，请求出最少需要调整多少根电线杆的高度、

输入n+1行
第一行整数n表示电线杆数
剩下n行表示电线杆的高度
10
6 3 4 3 3 4 5 5 4 6
输出2
注意是存在,至少调整的数量、

# 就是标记从左到右不满足非递减的数，标记从右到左不满足非递减的数，然后统计i处左右两的不满足条件的标记数，最后取最小
class Solution:
    def min_adjustment(self, n: int, hight: 'list[int]'):
        dec, inc = [0] * n, [0] * n
        def adjusted():
            # mark non-decreasing Numbers
            low = hight[0]
            for i in range(1,n):  # from left to right
                if hight[i] > low:
                    dec[i] = 1  # need be adjusted
                else:
                    low = hight[i]  # update the value of low
            # mark non-increasing Numbers
            low = hight[-1]
            for j in range(n-2, -1,-1):  # from right to left
                if hight[j] > low:
                    inc[j] = 1
                else:
                    low = hight[j]

        adjusted()

        res = [0] * n
        # Suppose i meets the conditions
        for i in range(n):
            res[i] = sum(dec[:i]) + sum(inc[i+1:])
        # return the smallest number
        return min(res)

if __name__=='__main__':
    s = Solution()
    n = int(input().strip())
    hight = list(map(int,input().strip().split()))
    print(s.min_adjustment(n, hight))

标签：__,adjusted,笔试,电线杆,int,编程,hight,low,乐鑫
来源： https://blog.csdn.net/hushaoqiqimingxing/article/details/98245567