最大连续子数组问题

 1 //暴力法
 2 int MaxSubArry(int *a, int n)
 3 {
 4       int maxsum = a[0];
 5       int currsum;
 6       for(int i = 0; i < n; ++i)
 7       {
 8           for(int j = i; j < n; ++j)
 9           {
10               currsum = 0;
11               for(int k = i; k < j; ++k)
12               {
13                   currsum += a[k];
14               }
15               if(currsum >= maxsum)
16               {
17                   maxsum = currsum;
18               }
19           }
20       }
21       return maxsum;
22 }

 //分治法
 23 int MaxSubArry(int *a, int from, int to)
 24 {
 25     //递归基
 26     if(from == to)
 27         return a[from];
 28     int middle = (from + to)/2;
 29     int m1 = MaxSubArry(a, from, middle);
 30     int m2 = MaxSubArry(a, middle + 1, to);
 31     int i, left = a[middle], now = a[middle];
 32     for(i = middle - 1; i > from; --i)
 33     {
 34         now += a[i];
 35         if(now > left)
 36         {
 37             left = now;
 38         }
 39     }
 40     int right = a[middle + 1];
 41     now = a[middle + 1];
 42     for(i = middle + 2; i < to; ++i)
 43     {
 44         now += a[i];
 45         if(now > right)
 46         {
 47             right = now;
 48         }
 49     }
 50     int m3 = left + right;
 51     return Max(m1, m2, m3);//求三者最大那个
 52 }

 //动态规划
 //记S[i]为以a[i]结尾的数组中和最大的子数组
 21 int MaxSubArry(int *a, int *s, int n)
 22 {
 23     int result = a[0];
 24     int sum = a[0];
 25     s[0] = a[0];
 26     for(int i = 1; i < n; ++i)
 27     {
 28         if(sum > 0)
 29         {
 30             sum += a[i];
 31             s[i] = sum;
 32         }
 33         else{
 34             result = sum + a[i];
 35             if(result > a[i])
 36             {
 37                 s[i] = result;
 38                 sum = result;
 39             }else
 40             {
 41                 s[i] = a[i];
 42                 sum = a[i];
 43             }
 44         }
 45     }
 46     return Max(s, n); //求s中最大的那个元素
 47 }