42. Trapping Rain Water (JAVA)

最新推荐文章于 2022-10-27 17:31:32 发布

转载最新推荐文章于 2022-10-27 17:31:32 发布 · 149 阅读

0 ·

CC 4.0 BY-SA版权

原文链接：http://www.cnblogs.com/qionglouyuyu/p/10815989.html

文章标签：

#java

博客围绕计算柱状图积水问题展开，给定非负整数数组代表柱状图，要计算下雨后能困住的水量。示例数组为[0,1,0,2,1,0,1,3,2,1,2,1]，可困住6个单位水量。还给出了计算方法，先找最高bar，再从左右分别计算面积，时间复杂度为O(n)。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Given n non-negative integers representing an elevation map where the width of each bar is 1, compute how much water it is able to trap after raining.

The above elevation map 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. Thanks Marcos for contributing this image!

Example:

Input: [0,1,0,2,1,0,1,3,2,1,2,1]
Output: 6

首先找到最高的bar，然后从左到这个最高的bar，依次计算面积；再从右到这个最高的bar，依次计算面积。时间复杂度n*2 = O(n)

class Solution {
    public int trap(int[] height) {
        int ret = 0;
        int curIndex = 0; //index of highest 
        int topIndex = 0;
        for(int i = 1; i < height.length; i++){
            if(height[i] > height[topIndex]) topIndex = i;
        }
        
        for(int i = 0; i < topIndex; i++){
            if(height[i] > height[curIndex]){
                curIndex = i;   
            }
            else{
                ret += (height[curIndex]-height[i]);
            }
        }
        
        curIndex = height.length-1;
        for(int i = height.length-2; i > topIndex; i--){
            if(height[i] > height[curIndex]){
                curIndex = i;   
            }
            else{
                ret += (height[curIndex]-height[i]);
            }
        }
        
        return ret;
    }
}