B. Splitting A Linked List (25)

時間限制 400 ms
內存限制 65536 kB
代碼長度限制 8000 B
判題程序 Standard 作者 CHEN, Yue

Given a singly linked list, you are supposed to rearrange its elements so that all the negative values appear before all of the non-negatives, and all the values in [0, K] appear before all those greater than K. The order of the elements inside each class must not be changed. For example, given the list being 18→7→-4→0→5→-6→10→11→-2 and K being 10, you must output -4→-6→-2→7→0→5→10→18→11.

Input Specification:

Each input file contains one test case. For each case, the first line contains the address of the first node, a positive N (<= 10⁵) which is the total number of nodes, and a positive K (<=1000). 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 Data Next

where Address is the position of the node, Data is an integer in [-10⁵, 10⁵], and Next is the position of the next node. It is guaranteed that the list is not empty.

Output Specification:

For each case, output in order (from beginning to the end of the list) the resulting linked list. Each node occupies a line, and is printed in the same format as in the input.

Sample Input: 00100 9 10 23333 10 27777 00000 0 99999 00100 18 12309 68237 -6 23333 33218 -4 00000 48652 -2 -1 99999 5 68237 27777 11 48652 12309 7 33218 Sample Output: 33218 -4 68237 68237 -6 48652 48652 -2 12309 12309 7 00000 00000 0 99999 99999 5 23333 23333 10 00100 00100 18 27777 27777 11 -1

題目大意：給一個鏈表和K，遍歷鏈表后將<0的結點先輸出，再將0～k區間的結點輸出，最后輸出>k的結點

分析：將結點用list[10000]保存，list為node類型，node中保存結點的值value和它的next地址。list的下標就是結點的地址。將<0、0～k、>k三部分的結點地址分別保存在v[0]、v[1]、v[2]中，最后將vector中的值依次輸出即可～

代碼：

#include<cstdio> #include<iostream> #include<vector> using namespace std;vector<int> v[3];struct node {int data;int next; }list[100000];int main() {int start,n,k,a;scanf("%d%d%d",&start,&n,&k);for(int i=0; i<n; i++){scanf("%d",&a);scanf("%d%d",&list[a].data,&list[a].next);}int p=start;while(p!=-1){if(list[p].data<0)v[0].push_back(p);else if(list[p].data>=0&&list[p].data<=k)v[1].push_back(p);elsev[2].push_back(p);p=list[p].next;}int flag=0;for(int i=0; i<3; i++)for(int j=0; j<v[i].size(); j++){if(flag==0){printf("%05d %d ",v[i][j],list[v[i][j]].data);flag=1;}else{printf("%05d\n%05d %d ",v[i][j],v[i][j],list[v[i][j]].data);}}printf("-1\n");return 0; }

---

與50位技術專家面對面20年技術見證，附贈技術全景圖

總結

以上是生活随笔為你收集整理的2017-9-17pat甲级 B的全部內容，希望文章能夠幫你解決所遇到的問題。

如果覺得生活随笔網站內容還不錯，歡迎將生活随笔推薦給好友。