从前向后遍历，用数组记录到当前位置之前最大的高度，从后向前记录最大的bar，两者中的小数-bar高度就是当前bar的存水量

class Solution(object):
    def trap(self, height):
        """
        :type height: List[int]
        :rtype: int
        """
        if len(height) == 0 or len(height) == 1 or len(height) == 2:
            return 0
        leftbar = [0 for i in range(len(height))]
        rightbar = height[-1]
        max_left = height[0]
        for i in range(len(height)):
            leftbar[i] = max(max_left,height[i])
            max_left = leftbar[i]
        sum_water = 0
        for i in range(len(height)-1,-1,-1):
            rightbar = max(rightbar,height[i])
            sum_water += min(leftbar[i],rightbar) - height[i]
        return sum_water

C++中 stack.peek()和stack.pop()都返回栈顶元素，区别在于前者不改变栈，后者删除了栈顶元素
思路如果栈为空或者当前bar长度小于栈顶所存坐标的长度时，将坐标加入栈
否则，当栈不为空，而且当前bar长度大于栈顶元素所存坐标长度时，栈pop（）min(当前bar长度，此时栈顶元素）-pop出的长度作为容器高度，当前bar的坐标-栈顶元素坐标-1，作为宽度，乘积计算此时的盛水量。循环此过程，直到不满足条件
当前坐标进栈

class Solution(object):
    def trap(self, height):
        """
        :type height: List[int]
        :rtype: int
        """
        stack = []
        sum_water = 0
        for i in range(len(height)):
            if len(stack) == 0 or height[i] < height[stack[-1]]:
                stack.append(i)
            else:
                while len(stack) > 0 and height[i] > height[stack[-1]]:
                    pre = stack.pop()
                    if len(stack) >0:
                        sum_water += (min(height[i],height[stack[-1]]) - height[pre]) * (i-stack[-1]-1)
            stack.append(i)
        return sum_water