1133 Splitting A Linked List
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 (≤103). 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
题意
第一行给定链表的起始地址 h、总结点数 n 和常量 k 。
要求将该链表划分成三个部分，将负数放在链表的最前面，正数放到负数的后面，并且数值范围在 [0,k] 的元素放在正数的前面，而大于 k 的元素放在正数的后面。
思路
具体思路如下：
1.先将结点信息存入数组当中，方便后续的调用。
2.准备三个数组 a,b,c ，分别存放负数元素的地址、[0,k] 之间的正数和大于 k 的正数，然后从前往后遍历链表，将元素地址放入对应的数组中。
3.将三个数组合并成一个数组，然后输出最终结果即可。
代码
#include<bits/stdc++.h> using namespace std; const int N = 100010; int h, n, k; int e[N], ne[N]; int main() { //输入每个结点的信息 cin >> h >> n >> k; for (int i = 0; i < n; i++) { int address, data, next; scanf("%d%d%d", &address, &data, &next); e[address] = data, ne[address] = next; } //对结点进行分类 vector<int> a, b, c; for (int i = h; i != -1; i = ne[i]) { int d = e[i]; if (d < 0) a.push_back(i); else if (d <= k) b.push_back(i); else c.push_back(i); } //将划分后的链表合并到一起 a.insert(a.end(), b.begin(), b.end()); a.insert(a.end(), c.begin(), c.end()); //输出最终链表 for (int i = 0; i < a.size(); i++) { printf("%05d %d ", a[i], e[a[i]]); if (i + 1 == a.size()) puts("-1"); else printf("%05d\n", a[i + 1]); } return 0; }
