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)); } }
LeetCode - Trapping Rain Water
原文：http://www.cnblogs.com/shuaiwhu/p/5075536.html