in LintCode Algorithm Array ~ read.

Trapping Rain Water

LintCode-363.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.

Example: Given [0,1,0,2,1,0,1,3,2,1,2,1], return 6.

Challenge:

O(n) time and O(1) memory

O(n) time and O(n) memory is also acceptable.

public class Solution {  
    /**
     * @param heights: an array of integers
     * @return: a integer
     */
    public int trapRainWater(int[] heights) {
        if (heights == null || heights.length == 0) {
            return 0;
        }

        int[] left = new int[heights.length];
        left[0] = heights[0];

        for (int i = 1; i < heights.length; i++) {
            left[i] = Math.max(left[i - 1], heights[i]);
        }

        int[] right = new int[heights.length];
        right[heights.length - 1] = heights[heights.length - 1];

        for (int i = heights.length - 2; i >= 0; i--) {
            right[i] = Math.max(right[i + 1], heights[i]);
        }

        int sum = 0;
        for (int i = 0; i < heights.length; i++) {
            int min = Math.min(left[i], right[i]);

            if (min - heights[i] > 0) {
                sum += min - heights[i];
            }
        }

        return sum;
    }
}

Hope this helps,
Michael