PAT1133:Splitting A Linked List
1133. Splitting A Linked List (25)
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 33218Sample 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
思路
逻辑水题,将一个链表划分成三个区间。
代码
#include<iostream> #include<vector> using namespace std; class node { public: int val; int next; }; vector<node> nodes(100000); int main() { vector<vector<int>> res(3); int start,N,K; while(cin >> start >> N >> K) { for(int i = 0;i < N;i++) { int tmp; cin >> tmp; cin >> nodes[tmp].val >> nodes[tmp].next; } while(start != -1) { if(nodes[start].val < 0) res[0].push_back(start); else if(nodes[start].val >= 0 && nodes[start].val <= K) res[1].push_back(start); else res[2].push_back(start); start = nodes[start].next; } int cnt = 0; for(int i = 0;i < res.size();i++) { for(int j = 0;j < res[i].size();j++) { if(cnt++ == 0) { printf("%05d %d ",res[i][j],nodes[res[i][j]].val); } else { printf("%05d\n%05d %d ",res[i][j],res[i][j],nodes[res[i][j]].val); } } } printf("-1"); } }