一、题目
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 -1二、题目大意
给定链表，按照特定顺序排序。
三、考点
链表
四、注意
1、有的节点可能不在链表内；
2、输出之前整合在一起，可以减少很多判断。
五、代码
#include<iostream>
#include<vector>
#define N 100005
using namespace std;
struct node {
int add;
int val;
int next;
};
vector<node> vec(N), v, v1, v2, v0;
int main() {
//read
int add, n, k;
cin >> add >> n >> k;
for (int i = 0; i < n; ++i) {
int a;
node nd;
cin >> a >> nd.val >> nd.next;
nd.add = a;
vec[a] = nd;
}
//clear
while (add != -1) {
v.push_back(vec[add]);
add = vec[add].next;
}
//sort
for(int i=0;i<v.size();++i){
if (v[i].val < 0)
v0.push_back(v[i]);
else if(v[i].val <= k)
v1.push_back(v[i]);
else
v2.push_back(v[i]);
}
//merge
v.clear();
for (int i = 0; i < v0.size(); ++i)
v.push_back(v0[i]);
for (int i = 0; i < v1.size(); ++i)
v.push_back(v1[i]);
for (int i = 0; i < v2.size(); ++i)
v.push_back(v2[i]);
if (v.size() == 0) {
cout << -1 << endl;
return 0;
}
//output
for (int i = 0; i < v.size() - 1; ++i) {
printf("%05d %d %05d\n", v[i].add, v[i].val, v[i + 1].add);
}
printf("%05d %d -1\n", v[v.size() - 1].add, v[v.size() - 1].val);
system("pause");
return 0;
}