原题链接在这里：http://www.lintcode.com/en/problem/maximum-subarray-ii/

题目：

Given an array of integers, find two non-overlapping subarrays which have the largest sum.
The number in each subarray should be contiguous.
Return the largest sum.

 Notice

The subarray should contain at least one number

Example

For given [1, 3, -1, 2, -1, 2], the two subarrays are [1, 3] and [2, -1, 2] or [1, 3, -1, 2] and [2], they both have the largest sum 7.

题解：

是Maximum Subarray的进阶版. 

可以建立left 保存从左向右 时 到i的最大subarray. right保存从右向左时的到i的最大subarray.

最后res 始终用left[i] + right[i+1]来保持最大.

Time Complexity: O(n). Space: O(n).

AC Java:

 1 public class Solution {
 2     /**
 3      * @param nums: A list of integers
 4      * @return: An integer denotes the sum of max two non-overlapping subarrays
 5      */
 6     public int maxTwoSubArrays(ArrayList<Integer> nums) {
 7         if(nums == null || nums.size() == 0){
 8             return 0;
 9         }
10         
11         int n = nums.size();
12         int [] left = new int[n];
13         int local = 0;
14         int global = Integer.MIN_VALUE;
15         for(int i = 0; i<n; i++){
16             local = Math.max(local + nums.get(i), nums.get(i));
17             global = Math.max(global, local);
18             left[i] = global;
19         }
20         
21         int [] right = new int[n];
22         local = 0;
23         global = Integer.MIN_VALUE;
24         for(int i = n-1; i>=0; i--){
25             local = Math.max(local + nums.get(i), nums.get(i));
26             global = Math.max(global, local);
27             right[i] = global;
28         }
29         
30         int res = Integer.MIN_VALUE;
31         for(int i = 0; i<n-1; i++){
32             res = Math.max(res, left[i] + right[i+1]);
33         }
34         return res;
35     }
36 }