最大子列求和问题

算法一：暴力遍历法

#include <stdio.h>

int MaxSubseqSum2( int A[], int N )  
{   int ThisSum, MaxSum = 0;
    int i, j;
    for( i = 0; i < N; i++ ) { /* i是子列左端位置 */
          for( j = i; j < N; j++ ) { /* j是子列右端位置 */
              ThisSum = 0;
              for( k = i; k <= j; k++)
              ThisSum += A[k];        
                  if( ThisSum > MaxSum ) /* 如果刚得到的这个子列和更大 */
                            MaxSum = ThisSum;    /* 则更新结果 */
          } /* j循环结束 */    
     } /* i循环结束 */    
     return MaxSum;  
}
int main() {
	int k;
	scanf("%d", &k);
	int a[10000000] = {0};
	for (int i = 0 ; i < k; i++)
		scanf("%d", &a[i]);
	printf("%d\n", MaxSubseqSum2(a, k));
	return 0;
}

此时算法一的时间复杂度为O(N^3)

算法二：算法一的部分优化

#include <stdio.h>

int MaxSubseqSum2( int A[], int N )  
{   int ThisSum, MaxSum = 0;
    int i, j;
    for( i = 0; i < N; i++ ) { /* i是子列左端位置 */
          ThisSum = 0;  /* ThisSum是从A[i]到A[j]的子列和 */
          for( j = i; j < N; j++ ) { /* j是子列右端位置 */
                  ThisSum += A[j];        /*对于相同的i，不同的j，只要在j-1次循环的基础上累加1项即可*/ 
                  if( ThisSum > MaxSum ) /* 如果刚得到的这个子列和更大 */
                            MaxSum = ThisSum;    /* 则更新结果 */
          } /* j循环结束 */    
     } /* i循环结束 */    
     return MaxSum;  
}
int main() {
	int k;
	scanf("%d", &k);
	int a[10000000] = {0};
	for (int i = 0 ; i < k; i++)
		scanf("%d", &a[i]);
	printf("%d\n", MaxSubseqSum2(a, k));
	return 0;
}

此时算法二的时间复杂度为O(N^2)

算法三：分而治之——递归

#include <stdio.h>

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 main() {
	int k;
	scanf("%d", &k);
	int a[10000000] = {0};
	for (int i = 0 ; i < k; i++)
		scanf("%d", &a[i]);
	printf("%d\n", MaxSubseqSum3(a, k));
	return 0;
}

此时算法三的时间复杂度为O(N*logN)

算法四：在线处理法

#include <stdio.h>

using namespace std;
int MaxSubseqSum(int *a, int k);
int main(){
	int k = 0;
	scanf("%d", &k);
	int a[1000000]={0};
	for (int i = 0; i < k; i++)
		scanf("%d", &a[i]);
	printf("%d\n", MaxSubseqSum(a, k));
	return 0;
}
int MaxSubseqSum(int *a, int k){
	int MaxSum, ThisSum;
	MaxSum = ThisSum = 0;
	for (int i = 0; i < k; i++) {
		ThisSum += a[i];
		if (ThisSum > MaxSum)
			MaxSum = ThisSum;
		else if (ThisSum < 0)
			ThisSum = 0;
	}
	return MaxSum;
}

此时算法四的时间复杂度为O(N)