最大连续子列和问题(Java) 

 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 }

 

posted @ 2019-05-25 20:48  Huayra  阅读(311)  评论(0)    收藏  举报