42. Trapping Rain Water

Problem:

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!

题意比较清楚了，就是给出一个数组序列，数组里元素的值代表木板的长度，然后问这样的结构能存多少的水。所以我的想法是求出每根柱子的左右板的最大长度，然后取两个的最小值，最后减去当前板的长度就是某根柱子能存的水量，所以算法大概如下：

1.从左到右比较，找出一个max_left数组，里面分别代表每根柱子左边的最长长度。

2.从右到左比较，建立一样的max_right数组。

3.然后取max_left[i]和max_right[i]的最小值，减去height[i]当前长度，就是当前柱子的存水量。

Code：(LeetCode运行12ms)

class Solution {public:    int trap(vector<int>& height) {        int size = height.size();        int *max_left = new int[size]();        int *max_right = new int[size]();                //从1开始，第一个没有左边，最后一个没有右边        for (int i = 1; i < size; i++) {            max_left[i] = max(max_left[i - 1], height[i - 1]);            max_right[size - i - 1] = max(max_right[size - i], height[size - i]);        }                int sum = 0;        for (int i = 0; i < size; i++) {            //取决于最短的板            int mins = min(max_left[i], max_right[i]);            if (mins > height[i]) {                sum += (mins - height[i]);            }        }        delete []max_left;        delete []max_right;        return sum;    }};