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.

rainwatertrap.png

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.

reference: https://www.youtube.com/watch?v=8BHqSdwyODs

Solution：Two pointers Interative

思路： 左右指针向中过程中，更新leftmax, rightmax, 小的那个的位置的水量是可以确定的(leftmax - height[i]) or (rightmax - height[j])

Time Complexity: O(N) Space Complexity: O(1)

class Solution {
    public int trap(int[] height) {
        if(height == null) return 0;
        
        int result = 0;
        int i = 0;
        int j = height.length - 1;
        
        int leftmax = 0;
        int rightmax = 0;
        while(i <= j) {
            leftmax = Math.max(leftmax, height[i]);
            rightmax = Math.max(rightmax, height[j]);
            if(leftmax < rightmax) {
                // leftmax is smaller than rightmax, so the (leftmax-height[a]) water can be stored
                result += (leftmax - height[i]);
                i++;
            }
            else {
                result += (rightmax - height[j]);
                j--;
            }
        }
        return result;
    }
}

思路：
Time Complexity: O(N) Space Complexity: O(N)

Solution Code：

class Solution {
    public int trap(int[] height) {
        if(height == null) return 0;
        
        int result = 0;
        int i = 0;
        int j = height.length - 1;
        
        int leftmax = 0;
        int rightmax = 0;
        while(i <= j) {
            leftmax = Math.max(leftmax, height[i]);
            rightmax = Math.max(rightmax, height[j]);
            if(leftmax < rightmax) {
                // leftmax is smaller than rightmax, so the (leftmax-height[a]) water can be stored
                result += (leftmax - height[i]);
                i++;
            }
            else {
                result += (rightmax - height[j]);
                j--;
            }
        }
        return result;
    }
}