Eric Chen Mock Interview

 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.
 1,-2,3,-1
 1，3 A
 -1 B
 7

 public int maxDiffSubArrays(int[] nums) {
 // write your code here

 }

 Solution: 

 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)); // 常规算 dp 的数组
             lGlobalMax[i] = Math.max(lLocalMax, lGlobalMax[i-1]);     // 将 lLocalMax 的结果变成 不以 nums[i] 结尾的数组， 方便最后的 maxHead[0, i], minEnd[i+1, n-1] 的O（n） 遍历。
             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;
}