02-1. Reversing Linked List (25)

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本文介绍了一种不使用排序方法解决特定链表遍历问题的思路与实现过程，通过重新组织节点来达到目标输出格式，适用于节点数量不大于10万的情况。

忙活了将近整整一天，终于把这个题给搞定了，因为没有用到链表的功能，也没有因为提示说不能先排序，排序必挂而放弃尝试，最终走了无数的弯路，还是给解决了。。

#include <stdio.h>

typedef struct{
	int curAdd;
	int seqNum;
	int nextAdd;
	}Node;

int main()
{	
	int startAdd;
	int N, K;
	scanf("%d %d %d", &startAdd, &N, &K);
	Node a[100000];
	int curAddI, seqNumI, nextAddI;
	for(int i=0; i<N; i++)
	{
		scanf("%d %d %d", &curAddI, &seqNumI, &nextAddI);
		a[i].curAdd = curAddI;
		a[i].seqNum = seqNumI;
		a[i].nextAdd = nextAddI;
	}
	
	int j = 0;		//j is the order of the array from 0 to N-1.
	Node temp;
	int curNum = 0;
	int effectNode = 0;
	int s;
	while(startAdd != -1)
	{
		if(startAdd != a[j].curAdd)
		{
			j++;
		}
		else
		{
			effectNode++;
			//printf("effectNode = %d\n", effectNode);
			startAdd = a[j].nextAdd;
			temp = a[j];
			for(s=j-1;s>=curNum;s--)
			{
				a[s+1] = a[s];
			}
			a[curNum] = temp;
			curNum++;
			j = curNum;			
		}		
	}
	/*for(int nn=0; nn<effectNode; nn++)
	{
		if(a[nn].nextAdd == -1)
		printf("%05d %d %d\n",a[nn].curAdd, a[nn].seqNum, -1);
		else
		printf("%05d %d %05d\n",a[nn].curAdd, a[nn].seqNum, a[nn].nextAdd);
	}*/
	int value = K;
	int count = 0;
	int m = K-1;		//m+1 is the output number in one round.
	int total = effectNode;
	
	if(effectNode < K)
	{
		for(int mm = 0; mm<effectNode-1; mm++)
		{
			printf("%05d %d %05d\n", a[mm].curAdd, a[mm].seqNum, a[mm+1].curAdd);
		}
		printf("%05d %d %d\n",a[effectNode-1].curAdd, a[effectNode-1].seqNum, -1);
		return 0;
	}
	else if(effectNode == K)
		{
			for(; m>0; m--)
			{
				printf("%05d %d %05d\n", a[m].curAdd, a[m].seqNum, a[m-1].curAdd);				
			}
			printf("%05d %d %d\n", a[0].curAdd, a[0].seqNum, -1);
			return 0;
		}	
	else
	{
		while(effectNode>=value)
		{					
			if(total%K == 0)
			{
				for(; m>=0; m--)
				{
					if((m != 0))
					{
						printf("%05d %d %05d\n", a[m+count].curAdd, a[m+count].seqNum, a[m+count-1].curAdd);
					}
					else if((m == 0) && (count == (total-value)))
					{
						printf("%05d %d %d\n",a[m+count].curAdd, a[m+count].seqNum, -1);
						return 0;
					}					
					else 
					{
						printf("%05d %d %05d\n", a[m+count].curAdd, a[m+count].seqNum, a[count+2*value-1].curAdd);
					}				
				}
				m = K-1;
				effectNode -=value;
				count +=value;
			}
			else
			{
				for(; m>=0; m--)
				{
					if(m != 0)
					{
						printf("%05d %d %05d\n", a[m+count].curAdd, a[m+count].seqNum, a[m+count-1].curAdd);
					}
					else if((m == 0) && ((total-count)>2*value))
					{
						printf("%05d %d %05d\n", a[m+count].curAdd, a[m+count].seqNum, a[count+2*value-1].curAdd);
					}
					else 
					{
						printf("%05d %d %05d\n", a[m+count].curAdd, a[m+count].seqNum, a[count+value].curAdd);
					}
				}
				m = K-1;
				effectNode -=value;
				count +=value;
			}
		}				
		for(;count<total-1;count++)
		{		
				printf("%05d %d %05d\n",a[count].curAdd, a[count].seqNum, a[count+1].curAdd);
		}
		printf("%05d %d %d\n",a[total-1].curAdd, a[total-1].seqNum, -1);
	}
	return 0;
}