最大子数组和
Posted 阿森森森
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#include <iostream>
using namespace std;
int Find_Max_Crossing_SubAr(int A[], int low, int mid, int high,int *max_left,int *max_right)
int left_sum = -10000000;
int sum = 0;
for (int i = mid; i >= low; i --)
sum += A[i];
if (sum >left_sum)
left_sum = sum;
*max_left=i;
int right_sum = -100000000;
sum = 0;
for (int j = mid + 1; j <= high; j ++)
sum += A[j];
if (sum > right_sum)
right_sum = sum;
*max_right=j;
return left_sum + right_sum;
int Find_Maximum_SubAr(int A[], int low, int high,int *max_left,int *max_right)
int left_sum, right_sum, cross_sum;
if (high == low)
*max_left=low;
*max_right=high;
return A[low];
else
int mid = (low + high) / 2;
left_sum = Find_Maximum_SubAr(A, low, mid,max_left,max_right);
right_sum = Find_Maximum_SubAr(A, mid + 1, high,max_left,max_right);
cross_sum = Find_Max_Crossing_SubAr(A, low, mid, high,max_left,max_right);
if (left_sum >= right_sum && left_sum >= cross_sum)
return left_sum;
else if (right_sum >= left_sum && right_sum >= cross_sum)
return right_sum;
else
return cross_sum;
int main()
int A[100];
int n;
cout<<"Input the number of numbers:";
cin>>n;
for (int i = 0; i < n; i ++)
cin>>A[i];
int s,l;
cout<<"The max sum of the subarray is:"<<Find_Maximum_SubAr(A, 0, n - 1,&s,&l)<<endl;
s=s+1;l=l+1;
cout<<"from "<<s<<" to "<<l<<endl;
return 0;
该算法复杂度O(n*logn);
主要思想:最大子数组区间必定存在于数组中点左方(不包括中点处)或跨越中点或数组中点右方(不包括中点)的区域,再使用递归便可求出。
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