Trapping Rain Water

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.

For example, 
Given [0,1,0,2,1,0,1,3,2,1,2,1], return 6.

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!

来源： <https://leetcode.com/problems/trapping-rain-water/>

分析

空间复杂度O(n)

对于每个柱子，找到其左右两边最高的柱子，该柱子能容纳的面积就是 min(max_left,max_right) - height。所以

1. 从左往右扫描一遍，对于每个柱子，求取左边最大值；

2. 从右往左扫描一遍，对于每个柱子，求最大右值；

3. 再扫描一遍，把每个柱子的面积并累加。

class Solution {
public:
    int trap(vector<int>& height) {
        int length=height.size();
        int *max_left=new int[length]();
        int *max_right=new int[length]();
        
        for(int i=1;i<length;i++){
            max_left[i]=max(max_left[i-1],height[i-1]);
            max_right[length-i-1]=max(max_right[length-i],height[length-i]);
        }
        
        int sum=0;
        for (int i=0;i<length;i++){
            int h=min(max_left[i],max_right[i]);
            if(h>height[i])
                sum+=h-height[i];
        }
        delete[]max_left;
        delete[]max_right;
        return sum;
    }
};

也可以，该方法空间复杂度O（1）

1. 扫描一遍，找到最高的柱子，这个柱子将数组分为两半；

2. 处理左边一半；

3. 处理右边一半。

 

class Solution {
public:
    int trap(vector<int>& height) {
        int max = 0; // 最高的柱子，将数组分为两半
        int n=height.size();
        for (int i = 0; i < n; i++)
            if (height[i] > height[max]) max = i;
        int water = 0;
        for (int i=0,peak=0;i<max;i++){
            if(peak<height[i])peak=height[i];
            else
                water+=peak-height[i];
        }    
        for(int j=n-1,peak=0;j>max;j--){
            if(peak<height[j])peak=height[j];
            else
                water+=peak-height[j];
        }
        
        return water;       
    }
};

来自为知笔记(Wiz)