LeetCode 53. Maximum Subarray（最大的子数组）

 1 public class Solution 
 2 {
 3     public int maxSubArray(int[] nums) 
 4     {
 5         //    Solution 1:  O(n)
 6         // check param validation.
 7         if(nums == null || nums.length == 0)
 8             return 0;
 9         
10         int sum = 0;
11         int max = Integer.MIN_VALUE;
12  
13         // iterate nums array.
14         for (int i = 0; i < nums.length; i++) 
15         {
16             // choose a larger one between current number or (previous sum + current number).
17             sum = Math.max(nums[i], sum + nums[i]);
18             max = Math.max(max, sum);    // choose the larger max.
19         }
20  
21         return max;
22     }
23     
24     
25     
26 }

 1 public class Solution 
 2 {
 3     public int maxSubArray(int[] nums) 
 4     {
 5         int max_ending_here = 0;
 6         int max_so_far = Integer.MIN_VALUE;
 7         
 8         for(int i = 0; i < nums.length; i++)
 9         {  
10             if(max_ending_here < 0) 
11                  max_ending_here = 0;  
12             max_ending_here += nums[i];  
13             max_so_far = Math.max(max_so_far, max_ending_here);   
14         }  
15         return max_so_far; 
16     }
17     
18     
19     
20 }

 1 public class Solution 
 2 {
 3     public int maxSubArray(int[] nums) 
 4     {
 5         // Solution 3: Divide and Conquer. O(nlogn)
 6         if(nums == null || nums.length == 0)
 7             return 0;
 8         
 9         
10         return Max_Subarray_Sum(nums, 0, nums.length-1);
11     }
12     
13     public int Max_Subarray_Sum(int[] nums, int left, int right)
14     {
15         if(left == right)    // base case: meaning there is only one element.
16             return nums[left];
17         
18         int middle = (left + right) / 2;    // calculate the middle one.
19         
20         // recursively call Max_Subarray_Sum to go down to base case.
21         int left_mss = Max_Subarray_Sum(nums, left, middle);    
22         int right_mss = Max_Subarray_Sum(nums, middle+1, right);
23         
24         // set up leftSum, rightSum and sum.
25         int leftSum = Integer.MIN_VALUE;
26         int rightSum = Integer.MIN_VALUE;
27         int sum = 0;
28         
29         // calculate the maximum subarray sum for right half part.
30         for(int i=middle+1; i<= right; i++)
31         {
32             sum += nums[i];
33             rightSum = Integer.max(rightSum, sum);
34         }
35         
36         sum = 0;    // reset the sum to 0.
37         
38         // calculate the maximum subarray sum for left half part.
39         for(int i=middle; i>= left; i--)
40         {
41             sum += nums[i];
42             leftSum = Integer.max(leftSum, sum);
43         }
44         
45         // choose the max between left and right from down level.
46         int res = Integer.max(left_mss, right_mss);
47         // choose the max between res and middle range.
48         
49         return Integer.max(res, leftSum + rightSum);
50         
51     }
52     
53 }