1 public class test
2 {
3 public static void main(String[] args)
4 {
5 int[] array = new int[] {13,-3,-25,20,-3,-16,-23,18,20,-7,12,-5,-22,15,-4,7};
6 int max_sum = Integer.MIN_VALUE;
7
8 //method1: Brutal Force O(N^2)
9 for(int i = 0; i < array.length; i++)
10 {
11 int sum = 0;
12 for(int j = i; j < array.length; j++)
13 {
14 sum += array[j];
15 if(sum > max_sum)
16 max_sum = sum;
17 }
18 }
19 System.out.println("The maximum sum is: " + max_sum);
20
21 //method2: Divide and Conquer O(NlogN)
22 max_sum = Maximum_Subarray(array, 0, array.length - 1);
23 System.out.println("The maximum sum is: " + max_sum);
24
25 //method3: 在线处理 O(N)
26 int cur_sum = 0;
27 for(int i = 0; i < array.length; i++)
28 {
29 cur_sum += array[i];
30 if(cur_sum > max_sum)
31 max_sum = cur_sum;
32 else if(cur_sum < 0)
33 cur_sum = 0;
34 }
35 System.out.println("The maximum sum is: " + max_sum);
36
37 //method4: DP(Dynamic Programming) O(N)
38 int[] dp = new int[array.length];
39 int tmpMax = array[0];
40 dp[0] = array[0];
41 for(int i = 1; i < dp.length; i++)
42 {
43 tmpMax += array[i];
44 dp[i] = max(dp[i - 1], tmpMax);
45 if(tmpMax < 0)
46 tmpMax = 0;
47 }
48 System.out.println("The maximum sum is: " + dp[dp.length - 1]);
49 }
50
51 public static int Maximum_Subarray(int[] array, int left, int right)
52 {
53 if(left == right)
54 return array[left];
55
56 int mid = left + (right - left) / 2;
57 int left_max = Maximum_Subarray(array, left, mid); //左子列的最大子列和
58 int right_max = Maximum_Subarray(array, mid + 1, right); //右子列的最大子列和
59 int crossing_max = Maximum_Crossing_Subarray(array, left, right, mid);
60 return max(left_max, right_max, crossing_max);
61 }
62
63 public static int Maximum_Crossing_Subarray(int[] array, int left, int right, int mid)
64 {
65 int l_sum = 0, l_max = Integer.MIN_VALUE;
66 for(int i = mid; i >= left; i--)
67 {
68 l_sum += array[i];
69 if(l_sum > l_max)
70 l_max = l_sum;
71 }
72 int r_sum = 0, r_max = Integer.MIN_VALUE;
73 for(int i = mid + 1; i <= right; i++)
74 {
75 r_sum += array[i];
76 if(r_sum > r_max)
77 r_max = r_sum;
78 }
79 return l_max + r_max;
80 }
81 }
