42. Trapping Rain Water

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!

链接： http://leetcode.com/problems/trapping-rain-water/

4/16/2017

BB电面准备

21ms, 77%

不是自己做出来的。思路：找到当前位置左边最大的bar和右边最大的bar，分别在2个数组中。选择其中较小的一边作为高度，宽度为1，并且减去当前bar自己的面积。

注意的问题：

1. forward, backward每一个值其实是不含跟height[i]比较的，是左边（不含自己）与右边（不含自己）

2. 第22行需要判断volume大于0，为什么？比方说[1,2,1]到中间元素时curVolume = -1，不要加到结果中去。

public class Solution {
    public int trap(int[] height) {
        if (height.length <= 2) return 0;
        int[] forward = new int[height.length];
        int[] backward = new int[height.length];

        int leftBar = height[0];
        for (int i = 1; i < height.length; i++) {
            forward[i] = leftBar;
            leftBar = Math.max(height[i], leftBar);
        }
        int rightBar = height[height.length - 1];
        for (int i = height.length - 2; i >= 0; i--) {
            backward[i] = rightBar;
            rightBar = Math.max(height[i], rightBar);
        }

        int sumVolume = 0;
        int curVolume;
        for (int i = 0; i < height.length; i++) {
            curVolume = Math.min(forward[i], backward[i]) - height[i];
            if (curVolume > 0) sumVolume += curVolume;
        }
        return sumVolume;
    }
}

双指针解法, 23ms, 52%

参考：http://www.cnblogs.com/yrbbest/p/4436338.html

https://discuss.leetcode.com/topic/5125/sharing-my-simple-c-code-o-n-time-o-1-space

跟之前一种解法其实思路是大致相同的。首先判断在相反方向自己是否是短板，若是，则判断是否高于之前的最高值，若是则更新最高值，若不是则加上当前高度差。

双指针为了什么？

自己认为是为了保证更新自己方面，比如quicksort是不符合规矩的一边改变index，这里是短板的一边计算volume值。

public class Solution {
    public int trap(int[] height) {
        if (height.length <= 2) return 0;
        int lo = 0, hi = height.length - 1;
        int maxL = 0, maxR = 0;
        int volume = 0;
        
        while (lo <= hi) {
            // find the left most index could pair with right
            if (height[lo] <= height[hi]) {
                // if current height less than its left max bar, it's in a basin, add volume
                if (height[lo] < maxL) {
                    // height difference is volume
                    volume += maxL - height[lo];
                } else {
                    // upward, update maxL
                    maxL = height[lo];
                }
                lo++;
            } else {
                if (height[hi] < maxR) {
                    volume += maxR - height[hi];
                } else {
                    maxR = height[hi];
                }
                hi--;
            }
        }
        return volume;
    }
}

还有一个stack的做法，有时间需要研究一下

https://discuss.leetcode.com/topic/4939/a-stack-based-solution-for-reference-inspired-by-histogram

更多讨论：

https://discuss.leetcode.com/category/50/trapping-rain-water