Program/Algorithm_practice 1.2

Program/Algorithm_practice 1.2

利用 online/dynamic array 解决子列最大和问题

---

• key point
  scan the array by order
  nowsum saves the current answer
  maxsum always saving the max value
  and when nowsum is minus ,drop it ,
  cause a minus only reduces the following value

---

#include <stdio.h>
#include <stdlib.h>

int online (int list[],int N){  
//list ,length of list =N
	
	int nowsum=0,maxsum=0,i=0;
	while (i<=N-1)
	{
		nowsum+=list[i];
		if (nowsum<0) nowsum=0;//drop the minus
		else if (nowsum>maxsum) maxsum=nowsum;
		//maxsum always saving the max
		i++;
	}
	return maxsum ;
}

int main(){
	int N,i,result;
	printf("please input the length of list :\n");
	
	scanf("%d",&N);

	int *l; 
	//dynamic array
	
	l= (int*)malloc(sizeof(int)*N);

		if(l)
		{
		printf("input numbers divided by space:\n");
		for(i=0;i<N;i++) scanf("%d",l+i);
			printf("\n");
			printf("your list is :\n");
			  
		for(i=0;i<N;i++) printf("%d\t",l[i]);
			printf("\n");
		}
		else printf("error!");

        result =online(l,N);
		printf("\nthe result is %d",result);

		free(l);
		l=NULL;
}

• result

please input the length of list :
8
input numbers divided by space:
4 -3 5 -2 -1 2 6 -2

your list is :
4       -3      5       -2      -1      2       6       -2

the result is 11

• Time complexity
  $$O(n)$$