Leetcode-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.

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!

Analysis:

We scan the whole array twice. First: from head to end; Second: from end to start.

In every scan, we record the current bar for rain trapping. If the value of current column is less than the bar, then current store += bar-currentValueOfColumn; Else, current store is end, we then add the current store into the total store, and reset the bar as currentValueOfColumn.

Solution:

public class Solution {
    public int trap(int[] A) {
        if (A.length==0) return 0;
        if (A.length==1) return 0;
        int index = 1;
        int bar = A[0];
        int total = 0;
        int curStore = 0;
        boolean[] count = new boolean[A.length];
        
        //We need to record which column has been counted during the first scan. 
        for (int i=0;i<count.length;i++) count[i] = false;
        List<Integer> countList = new ArrayList<Integer>();
        
        while (index<A.length){
            int curBar = A[index];
            if (curBar<bar){
                curStore+= (bar-curBar);
                countList.add(index);
            } else {
                total += curStore;
                curStore = 0;
                bar = curBar;
                for (int i=0;i<countList.size();i++) count[countList.get(i)]=true;
                countList.clear();
            }
            index++;
        }

        bar = A[A.length-1];
        index = A.length-2;
        curStore = 0;
        while (index>=0){
            int curBar = A[index];
            if (curBar<bar && !count[index])
                curStore+= (bar-curBar);
            else if (curBar>bar){
                total += curStore;
                curStore = 0;
                bar = curBar;
            } 
            index--;
        }


        return total;      
    }
}

NOTE: We need to record which column has been counted in the first scan, we then ignor these column in the second scan.