[leetcode]42. Trapping Rain Water(Java)

https://leetcode.com/problems/trapping-rain-water/#/description

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 Marcosfor contributing this image!

Java code:

package go.jacob.day623;import org.junit.Test;public class Demo1 {@Testpublic void testName() throws Exception {int[] num = { 0, 1, 0, 2, 1, 0, 1, 3, 2, 1, 2, 1 };System.out.println(trap(num));}/* * Runtime: 22 ms * 参考自leetcode网友：@yuyibestman  */public int trap(int[] height) {if (height == null || height.length < 2)return 0;int leftMax = height[0], rightMax = height[height.length - 1], left = 0, right = height.length - 1;int sum = 0;while (left < right) {leftMax = Math.max(height[left], leftMax);rightMax = Math.max(height[right], rightMax);if (leftMax < rightMax) {sum += leftMax - height[left];left++;} else {sum += rightMax - height[right];right--;}}return sum;}/* * 用总面积减去空白面积，再减去边的面积：分别从两侧从中间搜索，找到最高的边停止 runtime:21ms */public int trap_1(int[] height) {if (height == null || height.length < 2)return 0;int max = height[0];int sumOfNum = 0;// 记录空白面积int additionalArea = 0;// 找到最高边和所有边的和for (int i = 0; i < height.length; i++) {if (max < height[i])max = height[i];sumOfNum += height[i];}int tmpLarge = height[0];int otherEdge = 0;// 从左边开始搜索for (int i = 0; i < height.length; i++) {if (tmpLarge < height[i]) {additionalArea += (i - otherEdge) * (max - tmpLarge);otherEdge = i;tmpLarge = height[i];}// 找到最高边，停止搜索if (height[i] == max)break;}tmpLarge = height[height.length - 1];otherEdge = height.length - 1;// 从右边开始搜索for (int i = height.length - 1; i >= 0; i--) {if (tmpLarge < height[i]) {additionalArea += (otherEdge - i) * (max - tmpLarge);tmpLarge = height[i];otherEdge = i;}// 找到最高边，停止搜索if (height[i] == max)break;}int area = height.length * max - sumOfNum - additionalArea;return area;}}