PAT（甲级）2021年冬季考试 7-2 Rank a Linked List

7-2 Rank a Linked List
分析：
题意比较清楚简单，创建一个链表结构，记录结点的下一个结点和位置即可。
A linked list of n nodes is stored in an array of n elements. Each element contains an integer data and a next pointer which is the array index of the next node. It is guaranteed that the given list is linear – that is, every node, except the first one, has a unique previous node; and every node, except the last one, has a unique next node.

Now let’s number these nodes in order, starting from the first node, by numbers from 1 to n. These numbers are called the ranks of the nodes. Your job is to list their ranks.

Input Specification:
Each input file contains one test case. For each case, the first line gives a positive number n (≤105) which is the total number of nodes in the linked list. Then n numbers follow in the next line. The ith number (i=0,⋯,n−1) corresponds to next(i) of the ith element. The NULL pointer is represented by −1. The numbers in a line are separated by spaces.

Output Specification:
List n ranks in a line, where the ith number (i=0,⋯,n−1) corresponds to rank(i) of the ith element. The adjacent numbers in a line must be separated by exactly 1 space, and there must be no extra space at the beginning or the end of the line.

Sample Input:
5
3 -1 4 1 0
Sample Output:
3 5 1 4 2
Hint:
The given linked list is stored as 2->4->0->3->1->NULL. Hence the 0th element is ranked 3 since it is the 3rd node in the list; the 1st element is ranked 5 since it is the last (the 5th) node in the list; and so on so forth.
代码

#include<stdio.h>const int maxn = 100010;
struct Node{int next;int pos;
}node[maxn];int main(){int n, head, temp[maxn];bool vis[maxn] = {false};scanf("%d", &n);for(int i = 0; i < n; i++){scanf("%d", &node[i].next);if(node[i].next != -1){vis[node[i].next] = true;}temp[i] = node[i].next;}for(int i = 0; i < n; i++){if(vis[i] == false){head = i;break;}}int newnode = head;int count = 0;while(node[newnode].next != -1){node[newnode].pos = count++;newnode = node[newnode].next;}node[newnode].pos = count++;for(int i = 0; i < n; i++){if(temp[i] != -1){printf("%d", node[temp[i]].pos);}else{printf("%d", n);}if(i < n - 1){printf(" ");}}}

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PAT（甲级）2021年冬季考试 7-2 Rank a Linked List

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