拉丁方阵问题 (C语言循环链表的应用) ------- 算法笔记006

Dear 丶 2023-07-02 05:27 67阅读 0赞

## 问题解剖 ##

![在这里插入图片描述][watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzM1MzkyNzMz_size_16_color_FFFFFF_t_70]

## 实现代码 ##

#include <stdio.h>
    #include <stdlib.h>
    
    typedef struct node{ 
    	int data;
    	struct node *next;
    }Node,*List; 
    
    //创建一个规模为n*n的循环链表
    List createLoopList(List head,int n){ 
    	List p=head;
    	int i=1;
    	while(n--){ 
    		List new =(List)malloc(sizeof(Node));
    		new->data=i++;
    		p->next=new;
    		p=new;
    	}
    	p->next=head->next;
    	p=head->next;
    	free(head);
    	return p;
    }
    
    int main(int argc, char *argv[]) { 
    	int n=0;
    	List head=(List)malloc(sizeof(Node));
    	printf("------------------拉丁方阵问题------------------------\n");
    s:	printf("请输入方阵的规模x(n):");
    	scanf("%d",&n);
    	if(n>0&&n){ 
    		head=createLoopList(head,n);
    		List p=head;
    		int i,j;
    		for(i=0;i<n;i++){ 
    				p=head;
    				
    				//定位到第i+1个元素
    				for(j=0;j<i;j++)
    					p=p->next;
    					
    				//一次输出n个数据 
    		 		for(j=0;j<n;j++){ 
    		 			printf("%d\t",p->data);
    					p=p->next;	
    				 }
    		 		
    			printf("\n");
    		}
    	}else{ 
    		printf("小子！正经点!\n\n");
    		goto s;
    	} 
    	return 0;
    }

> 算法复杂度 O(n^2)

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