1: #include <stdio.h>
2: #include <stdlib.h>
3:
4: int sc=0;
5: int max3(int a,int b, int c)
6: {
7: sc++;
8: if(a>b)
9: {
10: return a>c?a:c;
11: }
12: else
13: {
14: return b>c?b:c;
15: }
16: }
17:
18: int max_sub_sum(const int v[], int left, int right)
19: {
20: int center;
21: int maxLeftHalfSum;
22: int maxRightHalfSum;
23: int leftHalfWithRightBorderSum;
24: int rightHalfWithLeftBorderSum;
25: int maxLeftHalfWithRightBorderSum;
26: int maxRightHalfWithLeftBorderSum;
27: int i;
28:
29: if(left==right)
30: {
31: return v[left]>0?v[left]:0;
32: }
33: center=(left+right)/2;
34:
35: maxLeftHalfWithRightBorderSum=leftHalfWithRightBorderSum=
36: maxRightHalfWithLeftBorderSum=rightHalfWithLeftBorderSum=0;
37: for(i=center;i>=left;i--)
38: {
39: leftHalfWithRightBorderSum+=v[i];
40: if(leftHalfWithRightBorderSum>maxLeftHalfWithRightBorderSum)
41: {
42: maxLeftHalfWithRightBorderSum=leftHalfWithRightBorderSum;
43: }
44: }
45:
46: for(i=center+1;i<=right;i++)
47: {
48: rightHalfWithLeftBorderSum+=v[i];
49: if(rightHalfWithLeftBorderSum>maxRightHalfWithLeftBorderSum)
50: {
51: maxRightHalfWithLeftBorderSum=rightHalfWithLeftBorderSum;
52: }
53: }
54:
55: return max3(max_sub_sum(v,left,center),max_sub_sum(v,center+1,right),
56: maxLeftHalfWithRightBorderSum+maxRightHalfWithLeftBorderSum);
57: }
58:
59: int main()
60: {
61: int v[]={1,-1,2,-2,5,-3,4,-4};
62: int r=max_sub_sum(v,0,7);
63: printf("the max sub sum is %d in step %d",r,sc);
64: return 0;
65: }
sc 用来统计max3执行的次数,这其实也是递归函数调用的次数,不妨假设N为2的整数次幂运算,即N=2k ,很明显依分治思路sc=1+21+22+…=20+21+22+…2k-1=2k-1=N-1.
通过简单推理可以知道sc其实就是把问题分治之后的所有子集的规模。
众所周知该算法的复杂度为O(NlogN),现在我们来看一下复杂度T(N)是怎麽求解出来的:
T(N)=N*20 +(N/2k-1)*21+(N/2k-2)*22+…(N/2)2k-1 =(k-1)*N //注意此处推导,是基于sc的推到结果,复杂度近似等于求和sc的每一次递归调用产生的37~53步的循环次数,还不知道如何输入求和符号。。汗
而k显然为logN,此处对数底为2,于是我们大致推导出算法的复杂度为N*(logN-1).