112 lines
2.5 KiB
C
112 lines
2.5 KiB
C
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/*1025 反转链表 （25 分）
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给定一个常数 K 以及一个单链表 L，请编写程序将 L 中每 K 个结点反转。例如：给定 L 为 1→2→3→4→5→6，K 为 3，则输出应该为 3→2→1→6→5→4；如果 K 为 4，则输出应该为 4→3→2→1→5→6，即最后不到 K 个元素不反转。
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输入格式：
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每个输入包含 1 个测试用例。每个测试用例第 1 行给出第 1 个结点的地址、结点总个数正整数 N (≤105)、以及正整数 K (≤N)，即要求反转的子链结点的个数。结点的地址是 5 位非负整数，NULL 地址用 −1 表示。
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接下来有 N 行，每行格式为：
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Address Data Next
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其中 Address 是结点地址，Data 是该结点保存的整数数据，Next 是下一结点的地址。
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输出格式：
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对每个测试用例，顺序输出反转后的链表，其上每个结点占一行，格式与输入相同
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输入样例：
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00100 6 4
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00000 4 99999
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00100 1 12309
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68237 6 -1
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33218 3 00000
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99999 5 68237
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12309 2 33218
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输出样例：
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00000 4 33218
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33218 3 12309
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12309 2 00100
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00100 1 99999
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99999 5 68237
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68237 6 -1
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*/
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//#pragma warning(disable:4996)
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#include<stdio.h>
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#include<stdlib.h>
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typedef struct LinkList
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{
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int Address;
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int Data;
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int Next;
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}linklist;
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int main()
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{
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linklist list[100000], *p1, *p2;
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int h_address, num, r_num, p_address, n_address, cnt = 0, tmp;
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p1 = (linklist*)malloc(sizeof(struct LinkList));
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p2 = (linklist*)malloc(sizeof(struct LinkList));
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scanf("%d %d %d", &h_address, &num, &r_num);
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getchar();
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for (int i = 0; i < num; i++)
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{
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int address, data, next;
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scanf("%d %d %d", &address, &data, &next);
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getchar();
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list[address].Address = address;
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list[address].Data = data;
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list[address].Next = next;
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}
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p1 = &list[h_address];
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for (;;)
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{
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cnt++;
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if (p1->Next == -1)
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break;
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p1 = &list[p1->Next];
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}
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tmp = h_address;
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while (cnt >= r_num)
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{
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p1 = &list[tmp];
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p_address = tmp;
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n_address = p1->Next;
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for (int i = 0; i < r_num; i++)
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{
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if (i)
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{
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n_address = p1->Next;
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p1->Next = p_address;
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p_address = p1->Address;
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}
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p1 = &list[n_address];
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}
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if (p2 != NULL)
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p2->Next = p_address;
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if (!i)
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h_address = p_address;
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list[tmp].Next = n_address;
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p2 = &list[tmp];
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tmp = p1->Address;
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cnt -= r_num;
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i++;
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}
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p1 = &list[h_address];
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for (;;)
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{
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if (p1->Next >= 0)
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printf("%05d %d %05d\n", p1->Address, p1->Data, p1->Next);
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else
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printf("%05d %d %d\n", p1->Address, p1->Data, p1->Next);
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if (p1->Next == -1)
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break;
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p1 = &list[p1->Next];
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}
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return 0;
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}
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