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.

[图片上传失败...(image-91728e-1513831694434)]
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!

一刷
题解：

1. 像Container with most water一样，设立一个lo = 0， hi = height.length - 1，使用双指针来夹逼遍历。
2. 设立一个res = 0， maxLeftBar = 0, maxRightBar = 0
3. 在 lo <= hi的条件下进行遍历
   1. 假如height[lo] <= height[hi]，或者height[lo] < height[hi]也可以， 这时候说明当前左边的height <= 右边的height。那么我们只需要考虑左边界和当前左边的height的差值，这个差值就是我们能容纳多少水

1. 在上述情况下，假如height[lo] >= maxLeftBar， 当前index的值 > maxLeftBar，那么我们不能接到水，我们要把maxLeftBar更新为height[lo]
2. 否则res += maxLeftBar - height[lo]，我们把这个差值加入到结果中
3. lo++
   2. 否则，左边的当前height > 右边的当前height，容易能盛多少水取决于右边的height以及maxRightBar的差值
4. 当height[hi] >= maxRightBar,我们更新maxRightBar = height[hi]
5. 否则，我们把结果 maxRightBar - height[hi]加入到res中
6. hi--

4. 最后返回结果res就可以了， 很巧妙。

Time Complexity - O(n)， Space Complexity - O(1)
还有stack 和dynamic programming的方法留给二刷三刷

public class Solution {
    public int trap(int[] height) {
        if(height == null || height.length == 0) return 0;
        int lo = 0, hi = height.length-1;
        int maxLeftBar = 0, maxRightBar = 0;
        int res = 0;
        while(lo<=hi){
            if(height[lo] <=height[hi]){
                if(height[lo] >=maxLeftBar) maxLeftBar = height[lo];
                else res += maxLeftBar - height[lo];
                lo++;
            }
            else{
               if(height[hi] >=maxRightBar) maxRightBar = height[hi];
                else res += maxRightBar - height[hi];
                hi--; 
            }
        }
        
        return res;
    }
}

二刷思路同上

class Solution {
    public int trap(int[] A) {
        int res = 0;
        int left = 0, right = A.length-1;
        int maxLeft = 0, maxRight = 0;
        while(left<=right){
            if(A[left]<A[right]){//regard the right as wall
                if(maxLeft<A[left]) maxLeft = A[left];
                else res += maxLeft -A[left];
                left++;
            }else{
                if(maxRight<A[right]) maxRight = A[right];
                else res += maxRight -A[right];
                right--;
            }
        }
        return res;
    }
}