02-线性结构3-Reversing Linked List-编程题

解题代码

#include<stdio.h>
#include<stdlib.h>
typedef struct Array array;
struct Array {
	int expon;
	int next;
};
int main()
{
	array a[100000];
	int i, j, head, N, reN, taddress, texpon, tnext, cnt, realN=0;
	scanf("%d %d %d", &head, &N, &reN);
	for (i = 0; i < N; i++) {
		scanf("%d %d %d", &taddress, &texpon, &tnext);
		a[taddress].expon = texpon;
		a[taddress].next = tnext;
	}
	int p = head;
	while (p!=-1) {
		realN++;
		p = a[p].next;
	}
	cnt = realN / reN;
	int  t, r, s, temp, flag=1;
	p = head, t = head, r = head, s = a[head].next;
	for (i = 0 ; i < cnt; i++) {
		for (j = 0; j < reN - 1;j++) {
			temp = a[s].next;
			a[s].next = r;
			r = s;
			s = temp;
		}
		if (flag) {
			head = r;
			flag = 0;
		}
		if (i == cnt - 1 && s != -1) {
			a[p].next = r;
			a[t].next = s;
		}
		else if (i == cnt - 1) {
			a[p].next = r;
			a[t].next = -1;
		}else if (i > 0 && i < cnt) {
			a[p].next = r;
			p = t;
		}
		t = s;
		r = s;
		s = a[s].next;
	}
	p = head;
	flag = 1;
	while (a[p].next!=-1) {
		if (flag) flag = 0;
		else printf("\n");
		printf("%05d %d %05d",p,a[p].expon,a[p].next);
		p = a[p].next;
	}
	if (flag) flag = 0;
	else printf("\n");
	printf("%05d %d %d", p, a[p].expon, a[p].next);
	return 0;
}

测试结果

问题整理

1.这题他大爷的。。。
2.		if (i == cnt - 1 && s != -1) {
		a[p].next = r;
		a[t].next = s;
	}
	else if (i == cnt - 1) {
		a[p].next = r;
		a[t].next = -1;
	}else if (i > 0 && i < cnt) {
		a[p].next = r;
		p = t;
	}//注意这段代码的判断逻辑。
3.注意会有多余结点不在链表上。