Trapping Raining Water 解答

Question

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.

Solution

referrence

Key to the solution is to know that for each point a[i], the max are is calculated by:

min(left,right) – a[i]

left is the maximum height before a[i], right is the maximum height after a[i].

Therefore, we can create two arrays to record left most height and right most height for each point. Time complexity O(n).

 public class Solution {
     public int trap(int[] height) {
         if (height == null || height.length < 1)
             return 0;
         int length = height.length;
         int[] leftMost = new int[length];
         int[] rightMost = new int[length];
         // First, find left biggest hight
         leftMost[0] = 0;
         for (int i = 1; i < length; i++)
             leftMost[i] = Math.max(leftMost[i - 1], height[i - 1]);

         // Then, find right biggest hight
         rightMost[length - 1] = 0;
         for (int i = length - 2; i >= 0; i--)
             rightMost[i] = Math.max(rightMost[i + 1], height[i + 1]);

         // Calculate sum
         int result = 0;
         for (int i = 0; i < length; i++) {
             int tmp = Math.min(leftMost[i], rightMost[i]) - height[i];
             if (tmp > 0)
                 result += tmp;
         }
         return result;
     }
 }