1133 Splitting A Linked List （25 分）

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 (≤10​5​​) which is the total number of nodes, and a positive K (≤10​3​​). 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 [−10​5​​,10​5​​], 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

 要注意刚开始输入的数据是无序的，需要一个tmp将其按链接顺序放在容器中，再进行3次遍历得到ans，输出即可。

代码如下：

---

#include<iostream>
#include<vector>
#include<string>
#include<algorithm>
#include<set>
#include<queue>
using namespace std;
struct Node {
	int ad, val, next;
}node[100010];
int hd, n, k;
int main() {
	cin >> hd >> n >> k;
	for (int i = 0; i < n; i++) {
		int a, b, c;
		scanf("%d%d%d", &a, &b, &c);
		node[a] = { a,b,c };
	}
	vector<Node> tmp, ans;
	for (int begin = hd; begin != -1; begin = node[begin].next) {
		tmp.push_back(node[begin]);
	}
	int l = tmp.size();
	for (int i = 0; i < l; i++) {
		if (tmp[i].val < 0) ans.push_back(tmp[i]);
	}
	for (int i = 0; i < l; i++) {
		if (tmp[i].val >= 0 && tmp[i].val <= k) ans.push_back(tmp[i]);
	}
	for (int i = 0; i < l; i++) {
		if (tmp[i].val > k) ans.push_back(tmp[i]);
	}
	for (int i = 0; i < l; i++) {
		if (i != l - 1) ans[i].next = ans[i + 1].ad;
		else ans[i].next = -1;
	}
	for (int i = 0; i < l; i++) {
		if (i != l - 1) printf("%05d %d %05d\n", ans[i].ad, ans[i].val, ans[i].next);
		else printf("%05d %d %d\n", ans[i].ad, ans[i].val, ans[i].next);
	}
	return 0;
}