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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.
The subarray should contain at least one number
Analysis:
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]+maxSubArray(p+1,i)}
we iterate p from i-1 to j-1, so we can record the max subarray we get at current p, this value can be used to calculate the max subarray from p-1 to i when p becomes p-1.
Solution:
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 }
Solution 2:
Use one dimension array.
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]; 12 for (int i=0;i<=len;i++) d[i] = 0; 13 14 for (int j=1;j<=k;j++) 15 for (int i=len;i>=j;i--){ 16 d[i] = Integer.MIN_VALUE; 17 int endMax = 0; 18 int max = Integer.MIN_VALUE; 19 for (int p=i-1;p>=j-1;p--){ 20 endMax = Math.max(nums.get(p), endMax+nums.get(p)); 21 max = Math.max(endMax,max); 22 if (d[i]<d[p]+max) 23 d[i] = d[p]+max; 24 } 25 } 26 27 return d[len]; 28 29 30 } 31 }
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原文地址：http://www.cnblogs.com/lishiblog/p/4183917.html
