[LeetCode] 42. Trapping Rain Water（收集雨水）

• Difficulty: Hard

• Related Topics: Array, Two Pointers, Stack

• Link: https://leetcode.com/problems/trapping-rain-water/

Description

Given n non-negative integers representing an elevation map where the width of each bar is 1, compute how much water it can trap after raining.
给定 n 个非负整数表示一个海拔地图，柱状图的单位宽度是 1，计算其在下雨后能收集多少雨水。

Examples

Example 1

Input: height = [0,1,0,2,1,0,1,3,2,1,2,1]
Output: 6
Explanation: The above elevation map (black section) is represented by array [0,1,0,2,1,0,1,3,2,1,2,1]. In this case, 6 units of rain water (blue section) are being trapped.

Example 2

Input: height = [4,2,0,3,2,5]
Output: 9

Constraints

• n == height.length
• 0 <= n <= 3e4
• 0 <= height[i] <= 1e5

Solution

这题乍看上去无从下手，实际理解了也不难。我们把关注点转移到每根『柱子』上，只要计算每根『柱子』能收集的雨水量，最后相加即可。大家都知道木桶原理吧，一根『柱子』的储水量取决于其左右两边较矮的那边，所以我们维护左右两边高度的最大值，以此计算每根柱子的储水量。代码如下：

import kotlin.math.max

class Solution {
    fun trap(height: IntArray): Int {
        var left = 0
        var right = height.lastIndex
        var leftMax = 0
        var rightMax = 0
        var result = 0

        while (left <= right) {
            leftMax = max(leftMax, height[left])
            rightMax = max(rightMax, height[right])

            if (leftMax < rightMax) {
                // 先不看右边，至少水不会从右边流走
                result += (leftMax - height[left])
                left++
            } else {
                // 右边同理，水不会从左边流走
                result += (rightMax - height[right])
                right--
            }
        }

        return result
    }
}