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.NoteThe subarray should contain at least one numberExampleGiven [-1,4,-2,3,-2,3],k=2, return 8Tags Expand 

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

 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 }

别人一个类似的方法，比我少一个loop，暂时没懂：

 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 }