leetcode845数组中的最长山脉

class Solution(object):
    def longestMountain(self, A):
        """
        :type A: List[int]
        :rtype: int
        """
        # 想法：一个用来记录左边山底，一个用来记录山顶，一个用来记录右边山底
        left=0
        top=left
        right=top
        maxlength=0
        while(left<len(A)-2):
            length=0
            # 首先找到左边山底
            for i in range(left,len(A)-1):
                if(A[i]<A[i+1]):
                    left=i
                    break
            # 找到山顶
            top=left
            while(top<len(A)-1 and A[top]<A[top+1]):
                top=top+1
            #找到右边山底
            right=top
            while(right<len(A)-1 and A[right]>A[right+1]):
                right=right+1
            # 山脉长度
            if(right>top and top>left):
                length = right - left+1
                left=right
            else:
                left=right+1
            if(length>maxlength):
                maxlength=length
            # 但是可能存在多个山脉
            
        return maxlength

偷懒了两天，这个代码效果不太好，明天再学习一下别人的解法，要考试了，加油复习啊，向着学弟冲啊

更简洁的写法：

class Solution(object):
    def longestMountain(self, A):
        """
        :type A: List[int]
        :rtype: int
        """
        end=base=0
        length=0
        while(base<len(A)-2):
            end=base
            if(end+1<len(A) and A[end]<A[end+1]):
                while(end+1<len(A) and A[end]<A[end+1]):
                    end+=1 # 找到山顶,end不断向前移动
                if(end+1<len(A) and A[end]>A[end+1]): # 要先判断有下山的过程
                    while(end+1<len(A) and A[end]>A[end+1]):
                        end+=1 # 找到右边山脚,end不断向前移动
                    length=max(length,end-base+1) # 写在有下山的条件里面
            base=max(end,base+1)
        return length

还有一种想法是，先找到山顶，往两边找山脚

class Solution(object):
    def longestMountain(self, A):
        """
        :type A: List[int]
        :rtype: int
        """
        length=0
        for i in range(1,len(A)-1):
            if(A[i]>A[i-1] and A[i]>A[i+1]):
                left=i-1
                right=i+1
                while(left>0 and A[left]>A[left-1]):
                    left-=1
                while(right<len(A)-1 and A[right]>A[right+1]):
                    right+=1
                length=max(length,right-left+1)
        return length

运行效率 3>2>1