42. Trapping Rain Water

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.
[外链图片转存失败,源站可能有防盗链机制,建议将图片保存下来直接上传(img-37ZzDROn-1571543148239)(http://www.leetcode.com/wp-content/uploads/2012/08/rainwatertrap.png)]

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!

思路：
从两边往中间搜索

class Solution {
    public int trap(int[] height) {
        int secHight = 0;  
        int left = 0;  
        int right = height.length-1;  
        int area = 0;  
        while (left < right){  
            if (height[left] <height[right]){  
                secHight = Math.max(height[left], secHight);  
                area += secHight-height[left];//计算当前格的能装雨水的容量  
                left++;  
            } else {  
                secHight = Math.max(height[right], secHight);  
                area += secHight-height[right];  
                right--;  
            }  
        }  
        return area;  
    }
}

class Solution {
    public int trap(int[] height) {
        int answer = 0;

        if(height.length > 2) {
            int[] maxLeft = new int[height.length],
                    maxRight = new int[height.length];
            maxLeft[0] = 0;
            maxRight[maxRight.length - 1] = 0;

            for (int left = 1; left < height.length; left++) {
                int right = height.length - 1 - left;
                maxLeft[left] = Math.max(maxLeft[left - 1], height[left - 1]);
                maxRight[right] = Math.max(maxRight[right + 1], height[right + 1]);
            }

            for(int i = 0; i < height.length; i++) {
                int difference = Math.min(maxLeft[i], maxRight[i]) - height[i];
                answer += difference > 0 ? difference : 0;
            }
        }

        return answer;
    }
}