最大子列和问题

最大子列和问题的几种算法比较

算法1时间复杂度：O（n³）

int MaxSubseqSum1(int A[],int N){
	int ThisSum,MaxSum = 0;
	int i,j,k;
	for (i=0;i<N;i++){
		for(j=i;j<N;j++){
			ThisSum = 0;
			for(k = i;k<=j;k++){/*子列从i开始，到j结束*/
				ThisSum += A[k];
				if(ThisSum>MaxSum){
					MaxSum = ThisSum;
				}
			}
		}
	}
	return MaxSum;
}

算法2时间复杂度：O(n²)

int MaxSubseqSum2(int A[],int N){
	int ThisSum,MaxSum = 0;
	int i,j;
	for (i=0;i<N;i++){
		// 子列起始i
		ThisSum = 0;
		for(j = i;j<N;j++){
			// 子列结束j
			ThisSum += A[j];
			if(ThisSum > MaxSum){
				MaxSum = ThisSum;
			}
		}
	}
	return MaxSum;
}

算法三时间复杂度：O(nlog(n))
T(n) = 2T(n/2)+ O(n) = 2(2T(n/2²）+Cn/2)+Cn=2^k ×T(n/2^k) +Ckn
令n = 2^k,T(n) = nT(1)+Cnlogn，当两个复杂度相加时，取大的，
该算法时间复杂度为O(nlog(n)）

int Max3(int A,int B,int C){
	return A>B?(A>C?A:C):(B>C?B:C);
}

// 分治法
int DevideAndConquer(int A[],int left,int right){
	int MaxLeftSum,MaxRightSum;
	int MaxLeftBorderSum,MaxRightBorderSum;

	int LeftBorderSum,RightBorderSum;
	int i,center;
	/*递归结束条件*/
	if(left == right){
		if (A[left]> 0) return A[left];
		else return 0;
	}

	center = (left + right)/2;
	// 递归获取两边子列的最大和
	MaxLeftSum = DevideAndConquer(A,left,center);
	MaxRightSum = DevideAndConquer(A,center+1,right);

	/*下面求跨越边界的子列最大和*/
	MaxLeftBorderSum = 0;
	LeftBorderSum = 0;
	for(i = center;i>=left;i--){
		LeftBorderSum += A[i];
		if(LeftBorderSum > MaxLeftBorderSum){
			MaxLeftBorderSum = LeftBorderSum;
		}
	}

	MaxRightBorderSum = 0;
	RightBorderSum = 0;
	for(i = center+1;i<=right;i++){
		RightBorderSum += A[i];
		if(RightBorderSum > MaxRightBorderSum){
			MaxRightBorderSum = RightBorderSum;
		}
	}
	// 获取最大子列和
	return Max3(MaxLeftSum,MaxRightSum,MaxLeftBorderSum+MaxRightBorderSum);
	
}
int MaxSubseqSum3(int A[],int N){
	// 与前两种算法接口保持一致
	return DevideAndConquer(A,0,N-1);
}

算法四时间复杂度：
O(n)

//在线处理
int MaxSubseqSum4(int A[], int N){
	int ThisSum,MaxSum;
	int i ;
	ThisSum = MaxSum =0;
	for(i=0;i<N;i++){
		ThisSum += A[i];
		if(ThisSum>MaxSum)
			MaxSum = ThisSum;
		else if(ThisSum<0){
		//当前和小于0，继续累加只会减小后面的值，弃用，从下一个重新累加
			ThisSum = 0;
		}
	}
	return MaxSum;
}

测试程序：调整S的大小来调整问题规模

// 最大子列和问题的四种算法分析及时间复杂度的统计

#include
#include
#include
#define S 1000
clock_t start,stop;
double duration;
int A[S];

// 算法1：
int MaxSubseqSum1(int A[],int N){
	int ThisSum,MaxSum = 0;
	int i,j,k;
	for (i=0;i<N;i++){
		for(j=i;j<N;j++){
			ThisSum = 0;
			for(k = i;k<=j;k++){
				ThisSum += A[k];
				if(ThisSum>MaxSum){
					MaxSum = ThisSum;
				}
			}
		}
	}
	return MaxSum;
}

// 算法二：
int MaxSubseqSum2(int A[],int N){
	int ThisSum,MaxSum = 0;
	int i,j;
	for (i=0;i<N;i++){
		// 子列起始i
		ThisSum = 0;
		for(j = i;j<N;j++){
			// 子列结束j
			ThisSum += A[j];
			if(ThisSum > MaxSum){
				MaxSum = ThisSum;
			}
		}
	}
	return MaxSum;
}

int Max3(int A,int B,int C){
	return A>B?(A>C?A:C):(B>C?B:C);
}

// 算法三：分治法
int DevideAndConquer(int A[],int left,int right){
	int MaxLeftSum,MaxRightSum;
	int MaxLeftBorderSum,MaxRightBorderSum;

	int LeftBorderSum,RightBorderSum;
	int i,center;
	
	if(left == right){
		if (A[left]> 0) return A[left];
		else return 0;
	}

	center = (left + right)/2;
	// 递归获取两边子列的最大和
	MaxLeftSum = DevideAndConquer(A,left,center);
	MaxRightSum = DevideAndConquer(A,center+1,right);

	/*下面求跨越边界的子列最大和*/
	MaxLeftBorderSum = 0;
	LeftBorderSum = 0;
	for(i = center;i>=left;i--){
		LeftBorderSum += A[i];
		if(LeftBorderSum > MaxLeftBorderSum){
			MaxLeftBorderSum = LeftBorderSum;
		}
	}

	MaxRightBorderSum = 0;
	RightBorderSum = 0;
	for(i = center+1;i<=right;i++){
		RightBorderSum += A[i];
		if(RightBorderSum > MaxRightBorderSum){
			MaxRightBorderSum = RightBorderSum;
		}
	}
	// 获取最大子列和
	return Max3(MaxLeftSum,MaxRightSum,MaxLeftBorderSum+MaxRightBorderSum);
	
}
int MaxSubseqSum3(int A[],int N){
	// 与前两种算法接口保持一致
	return DevideAndConquer(A,0,N-1);
}

// 算法四
int MaxSubseqSum4(int A[], int N){
	int ThisSum,MaxSum;
	int i ;
	ThisSum = MaxSum =0;
	for(i=0;i<N;i++){
		ThisSum += A[i];
		if(ThisSum>MaxSum)
			MaxSum = ThisSum;
		else if(ThisSum<0){
			ThisSum = 0;
		}
	}
	return MaxSum;
}
// 获取随机数存在数组A[S]中
void get_rand_A(){
	int i;
	for(i=1;i<=S;i++){
		srand((unsigned int)time(NULL)*i);
		A[i-1] = rand()%(101)-50; /*生成-50到50之间的随机数S个*/
		printf("%d\n",A[i-1]);
	}
}


int main(){	
	int i;
	get_rand_A();

	// 算法一的执行时间统计
	start = clock();
	MaxSubseqSum1(A,S);
	stop = clock();
	duration = ((double)(stop - start))/CLOCKS_PER_SEC;
	printf("tick = %f\n",(double)(stop - start));
	printf("duration = %6.2e\n",duration);

	// 算法二的执行时间统计
	start = clock();
	MaxSubseqSum2(A,S);
	stop = clock();
	duration = ((double)(stop - start))/CLOCKS_PER_SEC;
	printf("tick2 = %f\n",(double)(stop - start));
	printf("duration2 = %6.2e\n",duration);

	// 算法三的执行时间统计
	start = clock();
	MaxSubseqSum3(A,S);
	stop = clock();
	duration = ((double)(stop-start))/CLOCKS_PER_SEC;
	printf("tick3 = %f\n",(double)(stop - start));
	printf("duration3 = %6.2e\n",duration);

	// 算法四的执行时间统计
	start = clock();
	MaxSubseqSum4(A,S);
	stop = clock();
	duration = ((double)(stop - start))/CLOCKS_PER_SEC;
	printf("tick4 = %f\n",(double)(stop - start));
	printf("duration4 = %6.2e\n",duration);
	return 0;
}

当数列个数设为1000时，测试结果如下：tick1、2、3、4对应四种算法的时间。