最大子序列求和问题:
给定整数A1,A2……An(可能有负数),求其最大的自序列(为方便起见,如果所有整数均为负数,则最大子序列和为0)
说明加入指针数组是为了传出最大子序列数组
方法一:
int *MaxSubSeqenceSum1(const int A[], int N)
{
int Sum, MaxSum, i, j, k, m, n;
Sum = 0;
MaxSum = 0;
for (i = 0; i < N; i++)
{
for (j = i; j < N; j++)
{
Sum = 0;
for (k = i; k <= j; k++)
Sum += A[k];
if (Sum > MaxSum)
{
MaxSum = Sum;
m = i;
n = j;
}
}
}
int *p = (int*)malloc(sizeof(int*)*(3));
p[0] = m;
p[1] = n;
p[2] = MaxSum;
return p;
}
方法二:
int *MaxSubSeqenceSum2(const int A[], int N)
{
int Sum, MaxSum, i, j, m, n;
MaxSum = 0;
for (i = 0; i < N; i++)
{
Sum = 0;
for (j = i; j < N; j++)
{
Sum += A[j];
if (Sum > MaxSum)
{
MaxSum = Sum;
m = i;
n = j;
}
}
}
int *p = (int*)malloc(sizeof(int*) * 3);
p[0] = m;
p[1] = n;
p[2] = MaxSum;
return p;
}
方法三:
int *MaxSubSeqenceSum3(const int A[], int N)
{
int Sum, MaxSum, i,j, m, n;
Sum = MaxSum = 0;
j = m = n = 0; //m,n是分别计入数组最大时的下表引入j是为了得到m,n;
for (i = 0; i < N; i++)
{
Sum += A[i];
j++; //此次最大和的个数
if (Sum > MaxSum)
{
MaxSum = Sum;
n = i; //最大和的最大下标
m = n - j; //减去个数就是最大和的最小下表
}
else if (Sum < 0)
{
Sum = 0;
j = 0;
}
}
int *p = (int*)malloc(sizeof(int*) * 3);
if (p == NULL)
{
printf("Memory Allocated at: %x/n", p);
}
p[0] = m;
p[1] = n;
p[2] = MaxSum;
return p;
}
主函数:
int main()
{
int i;
int Array[10] = {5,-4,3,-8,2,9,-8,7,-1,6};
printf("待排序的数组:\n");
for (i = 0; i < 10; i++)
{
printf("%d ", Array[i]);
}
printf("\n");
int *q = NULL;
// q = MaxSubSeqenceSum1(Array, 10);//调用不同的函数
// q = MaxSubSeqenceSum2(Array, 10);
q = MaxSubSeqenceSum3(Array, 10);
printf("最大自序列和是:\n");
for (i = *q; i <= *(q+1); i++)
{
printf("%d ", Array[i]);
}
printf("\n");
printf("%d ", *(q+2));
free(q);
system("pause");
return 0;
}