最大和子序列问题

这个问题是算法导论的一个示例，为了讲解分治。

 1 //算法导论中的分治策略版本
 2 
 3 
 4 #include<iostream>
 5 using namespace std;
 6 int maxCrossSum(int a[], int begin, int mid, int end)
 7 {
 8     int sumLeft = a[mid];
 9     int sumNow=0;
10     for (int i = mid; i >= begin; --i)
11     {
12         sumNow += a[i];
13         if (sumNow > sumLeft)
14             sumLeft = sumNow;
15     }
16     int sumRight = a[mid + 1];
17     sumNow = 0;
18     for (int i = mid + 1; i <= end; ++i)
19     {
20         sumNow += a[i];
21         if (sumNow > sumRight)
22             sumRight = sumNow;
23     }
24     return sumLeft + sumRight;
25 }
26 int maxSubArray(int a[], int begin, int end)
27 {    
28     if (begin == end)
29         return a[begin]; 
30     else
31     {
32         int mid = (begin + end) / 2;
33         int leftSum = maxSubArray(a, begin, mid);
34         int rightSum = maxSubArray(a, mid + 1, end);
35         int crossSum = maxCrossSum(a, begin, mid, end);
36         int sum;
37         if (leftSum > rightSum)
38             sum = leftSum;
39         else sum = rightSum;
40         if (sum < crossSum)
41             sum = crossSum;
42         return sum;
43     }
44 }
45 int main()
46 {
47     const int SIZE = 13;
48     int a[SIZE] = { -3, -15, 20, -3, -16, -23, 18, 20, -9, 12, -5, -22, 15 };
49     int maxsubarray = maxSubArray(a, 0, 12);
50     cout << maxsubarray << endl;
51     system("pause");

这里提供一个更加简便的方法：

 1 更加高效的版本，无须递归，O（n）的时间复杂度
 2 #include<iostream>
 8 using namespace std;
 9 int maxSubArray(int a[], int begin, int end)
10 {    
11     int sum = 0;
12     int i = 0;
13     int maxSum = 0;
14     for (int i = begin; i <= end; ++i)
15     {    
16         sum += a[i];
17         if (sum > maxSum)
18             maxSum = sum;
19         //cout << maxSum << endl;
20         if (sum < 0)
21             sum = 0;
22     }
23     
24     return maxSum;
25 }
26 int main()
27 {
28     const int SIZE = 13;
29     int a[SIZE] = { -3, -15, 20, -3, -16, -23, 18, 20, -9, 12, -5, -22, 15 };
30     int maxsubarray = maxSubArray(a, 0, 12);
31     cout << maxsubarray << endl;
32     system("pause");
33 }

此处不得不说，后面的方法兼具大气魄，有远见等特点，方能如此简单便捷！