递归法在n个数中找个小于0的最大数.txt

来自「中过科学技术大学历年复试机试题」· 文本 代码 · 共 76 行

#include<stdio.h>
#include<stdlib.h>
#define N 10
 

//非递归法在N个数中找出小于0的最大数
void fin(int a[],int n){
  int i,min0max=0;
  for(i=0;i<N;i++)
	  if(a[i]<0){
		  min0max=a[i];
	      break;
	  }

  for(i=0;i<N;i++){
		  if(a[i]<0&&a[i]>min0max)min0max=a[i];
  }
	  
  printf("非递归法 the number required is %d\n",min0max);
   
}


//递归法在N个数中找个小于0的最大数
int f(int a[],int little,int high){
    int max1,max2,mid,min0max;
    if(little==high){
		if(a[little]<0)return a[little];
	    else return 0;   
	
	}
	
	if(little+1==high){
		if((a[little]*a[high])>0){
			if(a[little]>0&&a[high]>0)return 0;
			else if(a[little]<a[high])min0max=a[high];
			     else  min0max=a[little];
		}

	    if((a[little]*a[high])<=0){ 
		   if(a[little]<0)min0max=a[little];
	       else  min0max=a[high];     
        }
		return min0max;
	}
    
	else {
		  mid=(little+high)/2; 
          max1=f(a,little,mid);
          max2=f(a,mid+1,high);
		  if(max1*max2==0){
		       if(max1<0)return max1;
			   else return max2;
		  }
		  else if(max1<=max2)return max2;
		       else return max1;
	}
}


void main(){
  int a[N]={2,1,3,5,-4,6,-2,8,-6,-3};

  int min0max;
  
   min0max=f(a,0,N-1);
   


   printf("递归法  the number required is %d\n",min0max);
   fin(a,N);
   system("pause");
}