•方法一:双指针i,j满足i<j每次对[i,j]区间求和找出最大的值,复杂度是O(n^2)
#include<stdio.h>
#include<math.h>
#define N 1000000
int main()
{
int n, i, j;
int a[N];
int thissum, sum, tempi, tempj;
sum = -N;
scanf("%d", &n);
for (i = 0; i < n; i++)
{
scanf("%d", &a[i]);
}
for (i = 0; i < n; i++)
{
thissum = 0;
for (j = i; j < n; j++)
{
thissum += a[j];
if (thissum > sum)
{
sum = thissum;
tempi = i;
tempj = j;
}
}
}
printf("%d %d %d\n", sum, a[tempi], a[tempj]);
}
•方法二:分治法,分为左中右三部分,利用归并的方法求解O(nlogn)
int Max3( int A, int B, int C )
{ /* 返回3个整数中的最大值 */
return A > B ? A > C ? A : C : B > C ? B : C;
}
int DivideAndConquer( int List[], int left, int right )
{ /* 分治法求List[left]到List[right]的最大子列和 */
int MaxLeftSum, MaxRightSum; /* 存放左右子问题的解 */
int MaxLeftBorderSum, MaxRightBorderSum; /*存放跨分界线的结果*/
int LeftBorderSum, RightBorderSum;
int center, i;
if( left == right ) { /* 递归的终止条件,子列只有1个数字 */
if( List[left] > 0 ) return List[left];
else return 0;
}
/* 下面是"分"的过程 */
center = ( left + right ) / 2; /* 找到中分点 */
/* 递归求得两边子列的最大和 */
MaxLeftSum = DivideAndConquer( List, left, center );
MaxRightSum = DivideAndConquer( List, center+1, right );
/* 下面求跨分界线的最大子列和 */
MaxLeftBorderSum = 0; LeftBorderSum = 0;
for( i=center; i>=left; i-- ) { /* 从中线向左扫描 */
LeftBorderSum += List[i];
if( LeftBorderSum > MaxLeftBorderSum )
MaxLeftBorderSum = LeftBorderSum;
} /* 左边扫描结束 */
MaxRightBorderSum = 0; RightBorderSum = 0;
for( i=center+1; i<=right; i++ ) { /* 从中线向右扫描 */
RightBorderSum += List[i];
if( RightBorderSum > MaxRightBorderSum )
MaxRightBorderSum = RightBorderSum;
} /* 右边扫描结束 */
/* 下面返回"治"的结果 */
return Max3( MaxLeftSum, MaxRightSum, MaxLeftBorderSum + MaxRightBorderSum );
}
int MaxSubseqSum3( int List[], int N )
{ /* 保持与前2种算法相同的函数接口 */
return DivideAndConquer( List, 0, N-1 );
}
int MaxSubseqSum4( int A[], int N )
{ int ThisSum, MaxSum;
int i;
ThisSum = MaxSum = 0;
for( i = 0; i < N; i++ ) {
ThisSum += A[i]; /* 向右累加 */
if( ThisSum > MaxSum )
MaxSum = ThisSum; /* 发现更大和则更新当前结果 */
else if( ThisSum < 0 ) /* 如果当前子列和为负 */
ThisSum = 0; /* 则不可能使后面的部分和增大,抛弃之 */
}
return MaxSum;
}