转载标明出处：http://i.cnblogs.com/EditPosts.aspx?postid=4726782&update=1
暴力法:
1 // maxValue.cpp : 定义控制台应用程序的入口点。 2 // 3 4 #include "stdafx.h" 5 #include <iostream> 6 7 using namespace std; 8 9 int findMax(int *a,int low,int high,int &sub_left,int &sub_right,int &sub_sum) 10 { 11 if (low==high) 12 { 13 if(a[low]>0) 14 { 15 sub_left=low; 16 sub_right=high; 17 sub_sum=a[low]; 18 } 19 else sub_sum=0; 20 } 21 22 else 23 { 24 for(int i=0;i<=high;i++) 25 { 26 int tempSum=0; 27 for (int j=i;j<=high;j++) 28 { 29 tempSum+=a[j]; 30 if (tempSum>sub_sum) 31 { 32 sub_sum=tempSum; 33 sub_left=i; 34 sub_right=j; 35 } 36 } 37 } 38 } 39 return 0; 40 } 41 42 int _tmain(int argc, _TCHAR* argv[]) 43 { 44 int sub_left=0; 45 int sub_right=0; 46 int sub_sum=0; 47 int a[]={13,-3,-25,20,-3,-16,-23,18,20,-7,12,-5,-22,15,-4,7}; 48 findMax(a,0,15,sub_left,sub_right,sub_sum); 49 cout<<sub_left<<" "<<sub_right<<" "<<sub_sum<<endl; 50 system("pause"); 51 return 0; 52 }
时间复杂度：O(n*n);
分治法:
// maxSubArray.cpp : 定义控制台应用程序的入口点。 // #include "stdafx.h" #include <iostream> using namespace std; int Find_Max_cross_subArray(int *a,int low,int middle,int high,int &max_left,int &max_right,int &max_value) { int left_sum=-100; int sum=0; for (int i=middle;i>=low;i--) { sum+=a[i]; if (sum>left_sum) { left_sum=sum; max_left=i; } } int right_sum=-100; sum=0; for (int j=middle+1;j<=high;j++) { sum+=a[j]; if (sum>right_sum) { right_sum=sum; max_right=j; } } max_value=left_sum+right_sum; return 0; } int Find_Max_subArray(int *a,int low,int high,int &max_left,int &max_right,int &max_vlaue) { if (low==high) { if (a[low]>0) { max_vlaue=a[low]; max_left=low; max_right=high; } else max_vlaue=0; } else { int middle=(low+high)/2; int tem_left_low; int tem_left_high; int tem_left_sum; Find_Max_subArray(a,low,middle,tem_left_low,tem_left_high,tem_left_sum); int tem_right_low; int tem_right_high; int tem_right_sum; Find_Max_subArray(a,middle+1,high,tem_right_low,tem_right_high,tem_right_sum); int tem_cross_low; int tem_cross_high; int tem_cross_sum; Find_Max_cross_subArray(a,low,middle,high,tem_cross_low,tem_cross_high,tem_cross_sum); if (tem_left_sum>=tem_right_sum&&tem_left_sum>=tem_cross_sum) { max_left=tem_left_low; max_right=tem_left_high; max_vlaue=tem_left_sum; } else if (tem_right_sum>=tem_left_sum&&tem_right_sum>=tem_cross_sum) { max_left=tem_right_low; max_right=tem_right_high; max_vlaue=tem_right_sum; } else { max_left=tem_cross_low; max_right=tem_cross_high; max_vlaue=tem_cross_sum; } } return 0; } int _tmain(int argc, _TCHAR* argv[]) { int max_left=0; int max_right=0; int max_value=0; int a[]={13,-3,-25,20,-3,-16,-23,18,20,-7,12,-5,-22,15,-4,7}; Find_Max_subArray(a,0,15,max_left,max_right,max_value); cout<<max_left<<" "<<max_right<<" "<<max_value<<endl; system("pause"); return 0; }
递归式： T(n)=2*T(n/2)+Theta(n);n>1
T(n)=theta(1);
非递归线性算法：(4.1-5)
此方是看了我转载的那个伙伴的才有思路,和为负数，就跳过。
1 // Max_Value.cpp : 定义控制台应用程序的入口点。 2 // 3 4 #include "stdafx.h" 5 #include <iostream> 6 using namespace std; 7 8 int Max(int *a,int low,int high,int &sub_left,int &sub_right,int &sub_sum) 9 { 10 if (low==high) 11 { 12 if (a[low]>0) 13 { 14 sub_sum=a[low]; 15 sub_left=low; 16 sub_right=high; 17 } 18 else sub_sum=0; 19 } 20 int tempSum=0; 21 for (int i=low;i<=high;i++) 22 { 23 tempSum+=a[i]; 24 if (tempSum<=0) 25 { 26 tempSum=0; 27 sub_left=i+1; 28 } 29 if (tempSum>sub_sum) 30 { 31 sub_sum=tempSum; 32 sub_right=i; 33 } 34 } 35 return 0; 36 } 37 int _tmain(int argc, _TCHAR* argv[]) 38 { 39 int sub_left=0; 40 int sub_right=0; 41 int sub_sum=0; 42 int a[]={13,-3,-25,20,-3,-16,-23,18,20,-7,12,-5,-22,15,-4,7}; 43 Max(a,0,15,sub_left,sub_right,sub_sum); 44 cout<<sub_left<<" "<<sub_right<<" "<<sub_sum<<endl; 45 system("pause"); 46 return 0; 47 }
只扫描了一遍，时间复杂度为O(n);
以上方法输出结果都一样：7 10 43