leetcode 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.public class Solution { public static void main(String[] aargs){int[] input = {0,1,0,2,1,0,1,3,2,1,2,1};int result = trap(input);System.out.println(result);}public static int trap(int[] height) {       int result = 0;       if(height == null || height.length == 0){       return result;       }       int[] leftMaxHeight = new int[height.length];//从左至右的当前最大高度       int[] rightMaxHeight = new int[height.length];       int leftMax = 0;//左侧当前的最大高度       int rightMax = 0;       for (int i = 0; i < height.length; i++) {//从左至右扫描最大高度，并将其存入leftMaxHeight数组       leftMaxHeight[i] = leftMax;       leftMax = Math.max(height[i], leftMaxHeight[i]);       }       for (int i = height.length - 1; i >= 0; i--) {//从右至左扫描最大高度，并将其存入rightMaxHeight数组       rightMaxHeight[i] = rightMax;       rightMax = Math.max(height[i],rightMaxHeight[i]);       }       for(int i = 0;i<height.length;i++){       int min = Math.min(leftMaxHeight[i], rightMaxHeight[i]);//找到两个方向综合起来的最小高度       if(height[i]<min){//如果这个高度大于实际高度，则用这个高度减去实际高度就是本格可以装水的量       result = result+(min-height[i]);       }       }       return result;    }    }