Trapping Rain Water Leetcode

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!

 

 

 

这道题可能是第一道hard题目，一开始逻辑理了半天十分挫败，但是代码写出来，自己run了一次发现少了个等号，加了等号之后一遍submit就accepted了，好开心哈哈哈~也许是leetcode评难度偏低，感觉是medium的难度到不了hard。。。

这个题采用two pointers的方法，从左右到中间。主要的难点是判断什么时候left++什么时候right--，什么时候更新短板。

题中例子为例，当左边的木板不高于右边的木板的时候，就总是按照左边的高木板开始，往低于高木板的地方充水。如果遇到比这个高木板高的木板，但是低于右边的barrier，更新高木板，继续充水。

public class Solution {
    public int trap(int[] height) {
        if (height == null || height.length == 0) {
            return 0;
        }
        int left = 0;
        int right = height.length - 1;
        int count = 0;
        while (left < right) {
            int leftBarrier = height[left];
            while (left < right && height[left] <= height[right]) {
                if (height[left] < leftBarrier) {
                    count += leftBarrier - height[left];
                } else {
                    leftBarrier = height[left];
                }
                left++;
            }
            int rightBarrier = height[right];
            while (left < right && height[right] < height[left]) {
                if (height[right] < rightBarrier) {
                    count += rightBarrier - height[right];
                } else {
                    rightBarrier = height[right];
                }
                right--;
            }
        }
        return count;
    }
}

看了下top solution，思路是一样的，不过代码更简洁一些。

public class Solution {
    public int trap(int[] height) {
        if (height == null || height.length == 0) {
            return 0;
        }
        int left = 0;
        int right = height.length - 1;
        int count = 0;
        int leftBarrier = 0;
        int rightBarrier = 0;
        while (left < right) {
            leftBarrier = Math.max(leftBarrier, height[left]);
            rightBarrier = Math.max(rightBarrier, height[right]);
            if (leftBarrier < rightBarrier) {
                count += leftBarrier - height[left];
                left++;
            } else {
                count += rightBarrier - height[right];
                right--;
            }
        }
        return count;
    }
}