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 Tags 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 }