Lintcode: Maximum Subarray III

Given an array of integers and a number k, find k non-overlapping subarrays which have the largest sum.

The number in each subarray should be contiguous.

Return the largest sum.

Note
The subarray should contain at least one number

Example
Given [-1,4,-2,3,-2,3],k=2, return 8

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 1 public class Solution {
 2     /**
 3      * @param nums: A list of integers
 4      * @param k: An integer denote to find k non-overlapping subarrays
 5      * @return: An integer denote the sum of max k non-overlapping subarrays
 6      */
 7     public int maxSubArray(ArrayList<Integer> nums, int k) {
 8         // write your code
 9         if (nums.size() < k) return 0;
10         int len = nums.size();
11         
12         int[][] dp = new int[len+1][k+1];
13         
14         for (int i=1; i<=len; i++) {
15             for (int j=1; j<=k; j++) {
16                 if (i < j) {
17                     dp[i][j] = 0;
18                     continue;
19                 }
20                 dp[i][j] = Integer.MIN_VALUE;
21                 for (int p=j-1; p<=i-1; p++) {
22                     int local = nums.get(p);
23                     int global = local;
24                     for (int t=p+1; t<=i-1; t++) {
25                         local = Math.max(local+nums.get(t), nums.get(t));
26                         global = Math.max(local, global);
27                     }
28                     if (dp[i][j] < dp[p][j-1]+global) {
29                         dp[i][j] = dp[p][j-1]+global;
30                     }
31                 }
32             }
33         }
34         return dp[len][k];
35     }
36 }

 1 public class Solution {
 2     /**
 3      * @param nums: A list of integers
 4      * @param k: An integer denote to find k non-overlapping subarrays
 5      * @return: An integer denote the sum of max k non-overlapping subarrays
 6      */
 7     public int maxSubArray(ArrayList<Integer> nums, int k) {
 8         if (nums.size()<k) return 0;
 9         int len = nums.size();
10         //d[i][j]: select j subarrays from the first i elements, the max sum we can get.
11         int[][] d = new int[len+1][k+1];
12         for (int i=0;i<=len;i++) d[i][0] = 0;        
13         
14         for (int j=1;j<=k;j++)
15             for (int i=j;i<=len;i++){
16                 d[i][j] = Integer.MIN_VALUE;
17                 //Initial value of endMax and max should be taken care very very carefully.
18                 int endMax = 0;
19                 int max = Integer.MIN_VALUE;                
20                 for (int p=i-1;p>=j-1;p--){
21                     endMax = Math.max(nums.get(p), endMax+nums.get(p));
22                     max = Math.max(endMax,max);
23                     if (d[i][j]<d[p][j-1]+max)
24                         d[i][j] = d[p][j-1]+max;                    
25                 }
26             }
27 
28         return d[len][k];
29                     
30 
31     }
32 }