【Leetcode】Trapping Rain Water

標籤：

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：

用兩個數組，leftMax[i]表示左邊到i-1為之最高的高度，rightMax同上。

某位置能儲存的水量等於min{leftMax[i],rightMax[i]}-height[i]

演算法：

public int trap(int[] height) {      int water = 0;      int left_max[] = new int[height.length];      int right_max[] = new int[height.length];      for (int j = 1; j < height.length - 1; j++) {          left_max[j] = Math.max(left_max[j - 1], height[j - 1]);      }      for (int j = height.length - 2; j >=0; j--) {          right_max[j] = Math.max(right_max[j + 1], height[j + 1]);      }      for (int i = 1; i < height.length - 1; i++) {          int tmp = Math.min(left_max[i], right_max[i]) - height[i];          if (tmp > 0)              water += tmp;      }      return water;  }  

【Leetcode】Trapping Rain Water