42 Trapping Rain Water

Problem:

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!

Solutions:

public class Solution {
    public int trap(int[] height) {
        int left = 0;
        int leftIndex = -1;
        int water = 0;
        int inside = 0;
        HashSet<String> range = new HashSet<String>();
        for (int i = 0; i < height.length; i ++) {
            if (height[i] >= left) {
                range.add(leftIndex+","+i);
                water = water + left * (i - leftIndex - 1) - inside;
                left = height[i];
                leftIndex = i;
                inside = 0;

            }
            else {
                inside += height[i];
            }
        }
        int right = 0;
        int rightIndex = height.length;
        inside = 0;
        for (int i = height.length - 1; i >= 0; i --) {
            if (height[i] >= right) {
                if (!range.contains(i+","+rightIndex)) {
                    water = water + right * (rightIndex - i - 1) - inside;
                }
                right = height[i];
                rightIndex = i;
                inside = 0;
            }
            else {
                inside += height[i];
            }
        }
        return water;
    }
}

public class Solution {
    public int trap(int[] height) {
        if (height.length == 0) {
            return 0;
        }
        int[] left = new int[height.length];
        int[] right = new int[height.length];
        int max = height[0];
        left[0] = height[0];
        for (int i = 1; i < height.length; i ++) {
            if (height[i] > max) {
                max = height[i];
            }
            left[i] = max;
        }
        max = height[height.length - 1];
        right[height.length - 1] = height[height.length - 1];
        for (int i = height.length - 2; i >= 0; i --) {
            if (height[i] > max) {
                max = height[i];
            }
            right[i] = max;
        }
        int water = 0;
        for (int i = 0; i < height.length; i ++) {
            water +=Math.min(left[i], right[i]) - height[i];
        }
        return water;
    }
}