[Leetcode] 42. Trapping Rain Water 解题报告

题目：

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!

思路：

如果按照水平递增的方式计算可以收集到的雨水总和，思路还是比较复杂的（具体请见三维的雨水收集题目Leetcode 407）。但是对于二维情况有一种很好的线性解法：就是依次计算每个x坐标上可以收集到的雨水总和。对于位置x，它可以收集到的雨水取决于它左右最高挡板的最低值减去它本身的高度。而每个位置的左右最高挡板又可以采用贪心算法线性扫描得到并记录下来。这种解法的时间复杂度和空间复杂度都是O(n)，其中n是height的长度。

代码：

class Solution {public:    int trap(vector<int>& height) {        if(height.size() <= 2)            return 0;        int size = height.size();        vector<int> left(size, 0);        vector<int> right(size, 0);        for(int i = 1; i < size; ++i)            left[i] = max(left[i-1], height[i-1]);        for(int i = size - 2; i >= 0; --i)            right[i] = max(right[i+1], height[i+1]);        int sum = 0;        for(int i = 1; i < size - 1; ++i)        {            int min_height = min(left[i], right[i]);            sum += max(0, min_height - height[i]);        }        return sum;    }};