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.

思路：把数组分成两部分，可以从i和i+1(0<=  i < len-1)之间分开，a[0, i] a[i+1, len-1]，然后分别求两个子数组中的最大子段和，以及最小字段和，然后求差的最大值即可。

 1 public class Solution {

 2     /**

 3      * @param nums: A list of integers

 4      * @return: An integer indicate the value of maximum difference between two

 5      *          Subarrays

 6      */

 7     public int maxDiffSubArrays(ArrayList<Integer> nums) {

 8         // write your code

 9         if (nums==null || nums.size()==0) return 0;

10         int len = nums.size();

11         int[] lGlobalMax = new int[len];

12         int[] lGlobalMin = new int[len];

13         int lLocalMax = nums.get(0);

14         int lLocalMin = nums.get(0);

15         lGlobalMax[0] = lLocalMax;

16         lGlobalMin[0] = lLocalMin;

17         for (int i=1; i<len; i++) {

18             lLocalMax = Math.max(lLocalMax+nums.get(i), nums.get(i));

19             lGlobalMax[i] = Math.max(lLocalMax, lGlobalMax[i-1]);

20             

21             lLocalMin = Math.min(lLocalMin+nums.get(i), nums.get(i));

22             lGlobalMin[i] = Math.min(lLocalMin, lGlobalMin[i-1]);

23         }

24         

25         int[] rGlobalMax = new int[len];

26         int[] rGlobalMin = new int[len];

27         int rLocalMax = nums.get(len-1);

28         int rLocalMin = nums.get(len-1);

29         rGlobalMax[len-1] = rLocalMax;

30         rGlobalMin[len-1] = rLocalMin;

31         for (int i=len-2; i>=0; i--) {

32             rLocalMax = Math.max(rLocalMax+nums.get(i), nums.get(i));

33             rGlobalMax[i] = Math.max(rLocalMax, rGlobalMax[i+1]);

34             

35             rLocalMin = Math.min(rLocalMin+nums.get(i), nums.get(i));

36             rGlobalMin[i] = Math.min(rLocalMin, rGlobalMin[i+1]);

37         }

38         

39         int maxDiff = Integer.MIN_VALUE;

40         for (int i=0; i<len-1; i++) {

41             if (maxDiff < Math.abs(lGlobalMax[i]-rGlobalMin[i+1]))

42                 maxDiff = Math.abs(lGlobalMax[i]-rGlobalMin[i+1]);

43                 

44              

45             if (maxDiff < Math.abs(lGlobalMin[i]-rGlobalMax[i+1]))

46                 maxDiff = Math.abs(lGlobalMin[i]-rGlobalMax[i+1]);

47         }

48         return maxDiff;

49     }

50 }