1097. Deduplication on a Linked List (25)-PAT甲级真题

Given a singly linked list L with integer keys, you are supposed to remove the nodes with duplicated absolute values of the keys. That is, for each value K, only the first node of which the value or absolute value of its key equals K will be kept. At the mean time, all the removed nodes must be kept in a separate list. For example, given L being 21→-15→-15→-7→15, you must output 21→-15→-7, and the removed list -15→15.

Input Specification:

Each input file contains one test case. For each case, the first line contains the address of the first node, and a positive N (<= 105) which is the total number of nodes. The address of a node is a 5-digit nonnegative integer, and NULL is represented by -1.

Then N lines follow, each describes a node in the format:

Address Key Next

where Address is the position of the node, Key is an integer of which absolute value is no more than 104, and Next is the position of the next node.

Output Specification:

For each case, output the resulting linked list first, then the removed list. Each node occupies a line, and is printed in the same format as in the input.

Sample Input:
00100 5
99999 -7 87654
23854 -15 00000
87654 15 -1
00000 -15 99999
00100 21 23854
Sample Output:
00100 21 23854
23854 -15 99999
99999 -7 -1
00000 -15 87654
87654 15 -1

分析:用结构体数组存储这个链表,大小为maxn = 100000,node[i]表示地址为i的结点。在结构体中定义一个num变量,将num变量先初始化为2 * maxn。通过改变num变量的值最后sort排序来改变链表的顺序。
将没有删除的结点的num标记为cnt1,cnt1为当前没有删除的结点的个数;将需要删除的结点的num标记为maxn + cnt2,cnt2表示当前删除了的结点的个数,因为一开始初始化为了2 * maxn,所以我们可以通过对num排序达到:num = 0~maxn为不删除结点,num = maxn~2maxn为删除结点,num = 2maxn为无效结点


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