Lintcode: Maximum Subarray Difference

Given an array with integers.

Find two non-overlapping subarrays A and B, which |SUM(A) - SUM(B)| is the largest.

Return the largest difference.

Note
The subarray should contain at least one number

Example
For [1, 2, -3, 1], return 6

Challenge
O(n) time and O(n) space.

public class Solution {
/**
* @param nums: A list of integers
* @return: An integer indicate the value of maximum difference between two
*          Subarrays
*/
public int maxDiffSubArrays(ArrayList<Integer> nums) {
// write your code
if (nums==null || nums.size()==0) return 0;
int len = nums.size();
int[] lGlobalMax = new int[len];
int[] lGlobalMin = new int[len];
int lLocalMax = nums.get(0);
int lLocalMin = nums.get(0);
lGlobalMax[0] = lLocalMax;
lGlobalMin[0] = lLocalMin;
for (int i=1; i<len; i++) {
lLocalMax = Math.max(lLocalMax+nums.get(i), nums.get(i));
lGlobalMax[i] = Math.max(lLocalMax, lGlobalMax[i-1]);

lLocalMin = Math.min(lLocalMin+nums.get(i), nums.get(i));
lGlobalMin[i] = Math.min(lLocalMin, lGlobalMin[i-1]);
}

int[] rGlobalMax = new int[len];
int[] rGlobalMin = new int[len];
int rLocalMax = nums.get(len-1);
int rLocalMin = nums.get(len-1);
rGlobalMax[len-1] = rLocalMax;
rGlobalMin[len-1] = rLocalMin;
for (int i=len-2; i>=0; i--) {
rLocalMax = Math.max(rLocalMax+nums.get(i), nums.get(i));
rGlobalMax[i] = Math.max(rLocalMax, rGlobalMax[i+1]);

rLocalMin = Math.min(rLocalMin+nums.get(i), nums.get(i));
rGlobalMin[i] = Math.min(rLocalMin, rGlobalMin[i+1]);
}

int maxDiff = Integer.MIN_VALUE;
for (int i=0; i<len-1; i++) {
if (maxDiff < Math.abs(lGlobalMax[i]-rGlobalMin[i+1]))
maxDiff = Math.abs(lGlobalMax[i]-rGlobalMin[i+1]);

if (maxDiff < Math.abs(lGlobalMin[i]-rGlobalMax[i+1]))
maxDiff = Math.abs(lGlobalMin[i]-rGlobalMax[i+1]);
}
return maxDiff;
}
}