技術標籤：LeetCode300-javaleetcode

文章目錄

• 題目地址
• 題目描述
• 思路
• 題解

題目地址

https://leetcode-cn.com/problems/trapping-rain-water/

題目描述

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.

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 <= 3 * 10^4
• 0 <= height[i] <= 10^5

思路

對每個柱子上的積水，由其左右兩邊最高的柱子中較矮的柱子決定，因此可以計算每個柱子左右最高的柱子，然後計算其中矮的柱子和當前柱子的差。

題解

class Solution {
    public int trap(int[] height) {
        int res=0,i,j;
        int len = height.length;
        int[] left = new int[len];
        int[] right = new int[len];

        int lmax=0;
        for(i=0;i< height.
 
length;++i){
            left[i]=lmax;
            lmax = Math.max(lmax,height[i]);
        }
        int rmax = 0;
        for(j= height.length-1;j>=0;--j){
            right[j]=rmax;
            rmax = Math.max(rmax,height[j]);
        }
        
        for(i=0;i<len;i++){
            int h = Math.min(left[i],right[i]);
            if(h>height[i]) res+= h - height[i];
        }

        return res;
    }
}