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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DP. d[i][j] means the maximum sum we can get by selecting j subarrays from the first i elements.

d[i][j] = max{d[p][j-1]+maxSubArrayindexrange(p,i-1)}, with p in the range j-1<=p<=i-1

 public class Solution {
     /**
      * @param nums: A list of integers
      * @param k: An integer denote to find k non-overlapping subarrays
      * @return: An integer denote the sum of max k non-overlapping subarrays
      */
     public int maxSubArray(ArrayList<Integer> nums, int k) {
         // write your code
         if (nums.size() < k) return 0;
         int len = nums.size();

         int[][] dp = new int[len+1][k+1];

         for (int i=1; i<=len; i++) {
             for (int j=1; j<=k; j++) {
                 if (i < j) {
                     dp[i][j] = 0;
                     continue;
                 }
                 dp[i][j] = Integer.MIN_VALUE;
                 for (int p=j-1; p<=i-1; p++) {
                     int local = nums.get(p);
                     int global = local;
                     for (int t=p+1; t<=i-1; t++) {
                         local = Math.max(local+nums.get(t), nums.get(t));
                         global = Math.max(local, global);
                     }
                     if (dp[i][j] < dp[p][j-1]+global) {
                         dp[i][j] = dp[p][j-1]+global;
                     }
                 }
             }
         }
         return dp[len][k];
     }
 }

别人一个类似的方法，比我少一个loop，暂时没懂：

 public class Solution {
     /**
      * @param nums: A list of integers
      * @param k: An integer denote to find k non-overlapping subarrays
      * @return: An integer denote the sum of max k non-overlapping subarrays
      */
     public int maxSubArray(ArrayList<Integer> nums, int k) {
         if (nums.size()<k) return 0;
         int len = nums.size();
         //d[i][j]: select j subarrays from the first i elements, the max sum we can get.
         int[][] d = new int[len+1][k+1];
         for (int i=0;i<=len;i++) d[i][0] = 0;        

         for (int j=1;j<=k;j++)
             for (int i=j;i<=len;i++){
                 d[i][j] = Integer.MIN_VALUE;
                 //Initial value of endMax and max should be taken care very very carefully.
                 int endMax = 0;
                 int max = Integer.MIN_VALUE;
                 for (int p=i-1;p>=j-1;p--){
                     endMax = Math.max(nums.get(p), endMax+nums.get(p));
                     max = Math.max(endMax,max);
                     if (d[i][j]<d[p][j-1]+max)
                         d[i][j] = d[p][j-1]+max;
                 }
             }

         return d[len][k];

     }
 }