Trapping Rain Water

最新推荐文章于 2025-07-31 22:19:59 发布

kotic

最新推荐文章于 2025-07-31 22:19:59 发布

阅读量44

点赞数

CC 4.0 BY-SA版权

分类专栏： leetcode算法学习文章标签： python

本文链接：https://blog.youkuaiyun.com/kduduuuw/article/details/134576971

leetcode算法学习专栏收录该内容

71 篇文章

订阅专栏

文章介绍了如何使用两指针方法解决在给定海拔地图上计算降雨后每个位置被困水体积的问题，通过遍历地图并动态更新左右边界的最大高度来实现，时间复杂度为O(n)，空间复杂度为O(1)。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Intuition

The problem involves calculating the amount of water that can be trapped between the bars of an elevation map after raining. The idea is to use a two-pointer approach to determine the maximum height of bars on the left and right of each position and calculate the water trapped at each position.

Approach

The provided solution uses a two-pointer approach to iterate through the elevation map and calculate the trapped water at each position.

Initialize two pointers, left and right, at the beginning and end of the elevation map, respectively.
Initialize variables leftmax and rightmax to keep track of the maximum height of bars on the left and right, respectively. Set them to the height of the first and last bars.
Initialize a variable res to keep track of the total trapped water.
While the left pointer is less than the right pointer:
If leftmax is less than rightmax, move the left pointer to the right and update leftmax. Calculate and add the trapped water at the current position to res.
If rightmax is less than or equal to leftmax, move the right pointer to the left and update rightmax. Calculate and add the trapped water at the current position to res.
Repeat the process until the pointers meet.

Complexity

Time complexity:

The time complexity of this solution is O(n), where n is the length of the input array height. The two-pointer approach allows for a single pass through the elevation map.

Space complexity:

The space complexity is O(1), as the solution uses only a constant amount of space to store pointers and variables.

Code

class Solution:
    def trap(self, height: List[int]) -> int:
        if not height: return 0

        left , right = 0 , len(height) - 1
        leftmax,rightmax = height[left] , height[right]
        res=0

        while left < right :
            if leftmax < rightmax:
                left+=1
                leftmax=max(height[left],leftmax)
                res += leftmax - height[left]
            else:
                right-=1
                rightmax=max(height[right],rightmax)
                res += rightmax -height[right]
        
        return res