LeetCode

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!

思路：

1. 找到最高的点index

2. 计算(0, index)和(index, n-1)之间各点的面积

package area;

public class TrappingRainWater {

    public int trap(int[] height) {
        int n;
        if (height == null || (n = height.length) == 0) return 0;
        int area = 0;
        int maxHeight = height[0];
        int maxHeightIndex = 0;
        for (int i = 1; i < n; ++i) {
            if (height[i] > maxHeight) {
                maxHeight = height[i];
                maxHeightIndex = i;
            }
        }
        
        int leftMax = height[0];
        for (int i = 1; i < maxHeightIndex; ++i) {
            if (height[i] > leftMax)
                leftMax = height[i];
            area += (leftMax - height[i]);
        }
        
        int rightMax = height[n - 1];
        for (int i = n - 2; i > maxHeightIndex; --i) {
            if (height[i] > rightMax) 
                rightMax = height[i];
            area += rightMax - height[i];
        }
        
        return area;
    }
    
    public static void main(String[] args) {
        // TODO Auto-generated method stub
        int[] height = { /*0,1,0,2,1,0,1,3,2,1,2,1*/ 5,2,1,2,1,5 };
        TrappingRainWater t = new TrappingRainWater();
        System.out.println(t.trap(height));
    }

}