1133. Splitting A Linked List (25)
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:读入链表后遍历一次,将<0 <=k >k 分别存放,最后依次输出
注意:1.输入可能有无效节点,所以一定要遍历一次过滤
2.next是根据重新排列以后的顺序来定的
3.用数组模拟链表会比较方便
4.输出要固定5个数字,所以用printf会比较方便
5.最后一个-1 要和第4条区别开
6.考虑极端情况,例如没有<0 的时候 或者只有 <0 的时候
代码:
#include <iostream> #include <stdlib.h> #include <iomanip> #include <string.h> using namespace std; int nxt[100000]; int list[100000]; int minlist[100000]; int j = 0; int midlist[100000]; int p = 0; int maxlist[100000]; int q = 0; int main() { int add, k, num; cin >> add >> num >> k; int a, b, c; memset(nxt, -1, sizeof(int) * 100000); memset(list, -1, sizeof(int) * 100000); for (int i = 0; i < num; i++) { cin >> a >> b >> c; list[a] = b; nxt[a] = c; } a = add; while (a!=-1) { if (list[a] < 0) minlist[j++] = a; else if (list[a] >= 0 && list[a] <= k) midlist[p++] = a; else maxlist[q++] = a; a = nxt[a]; } int judge = 0; for (int i = 0; i < j; i++) { if (judge) cout << setw(5) << setfill('0') << minlist[i] << endl; cout << setw(5) << setfill('0') << minlist[i] << " " << list[minlist[i]] << " "; judge = 1; } for (int i = 0; i < p; i++) { if (judge) cout << setw(5) << setfill('0') << midlist[i] << endl; cout << setw(5) << setfill('0') << midlist[i] << " " << list[midlist[i]] << " "; judge = 1; } for (int i = 0; i < q; i++) { if (judge) cout << setw(5) << setfill('0') << maxlist[i] << endl; cout << setw(5) << setfill('0') << maxlist[i] << " " << list[maxlist[i]] << " "; judge = 1; } cout << -1 << endl; }
上面的代码有很多可以优化的地方,比如用print可以看起来精简很多,最后三个的输出可以用一个for来写,开头申请三个list太占空间用vector可以完美解决
感言:这题似乎是第二次做,然后还是折腾了半天,各种莫名奇妙的bug,还有各种考虑不周到。不过开心的是,这次的写的代码比上次好看多了 o(* ̄▽ ̄*)ブ
测试点:第一个测试点是样例
第二个测试点(估计)是复杂情况的样例,三种情况都输出
第三个测试点第四个测试点没有<0 的数据
第五个测试点(估计)有无效节点,需要过滤