[LeetCode]（面试题 17.21）直方图的水量

题目

给定一个直方图(也称柱状图)，假设有人从上面源源不断地倒水，最后直方图能存多少水量?直方图的宽度为 1。

上面是由数组 [0,1,0,2,1,0,1,3,2,1,2,1] 表示的直方图，在这种情况下，可以接 6 个单位的水（蓝色部分表示水）。

示例:

输入: [0,1,0,2,1,0,1,3,2,1,2,1]
输出: 6
           

解题思路

具体思路可参考 42. 接雨水

代码

解法一：动态规划

class Solution:
    def trap(self, height: List[int]) -> int:
        n = len(height)
        if n == 0:
            return 0
        leftmax = [0]*n
        rightmax = [0]*n
        left_max = 0
        right_max = 0
        res = 0
        for i in range(n):
            if height[i]>left_max:
                left_max = height[i]
            leftmax[i] = left_max
            if height[n-1-i]>right_max:
                right_max = height[n-1-i]
            rightmax[n-1-i] = right_max
        for i in range(n):
            res += min(leftmax[i], rightmax[i]) - height[i]
        return res
           

解法二：双指针

class Solution:
    def trap(self, height: List[int]) -> int:
        n = len(height)
        if n == 0:
            return 0
        leftmax = 0
        rightmax = 0
        left = 0
        right = n-1
        res = 0
        while left<=right:
            if leftmax<rightmax:
                leftmax = max(leftmax, height[left])
                res += leftmax - height[left]
                left += 1
            else:
                rightmax = max(rightmax, height[right])
                res += rightmax - height[right]
                right -= 1
        return res
           

解法三：单调栈

class Solution:
    def trap(self, height: List[int]) -> int:
        n = len(height)
        if n == 0:
            return 0
        stack = []
        res = 0
        for i in range(n):
            while stack and height[stack[-1]] < height[i]:
                bottomid = stack.pop()
                while stack and height[stack[-1]] == height[bottomid]:
                    stack.pop()
                if stack:
                    h = min(height[stack[-1]], height[i]) - height[bottomid]
                    w = i - stack[-1] - 1
                    res += h*w
            stack.append(i)
        return res