17年春季第二题 1133 Splitting A Linked List（25分）自古华山一条路

矫情吗；* 2023-07-04 14:12 6阅读 0赞

# 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

# 解释 #

建的两个vector好比左右手，左手倒右手，倒着倒着就行了。

（这道题的解法我是从liuchuo的博客那里学到的，发现用来解这种类型的链表题又快又好，谢谢她的工作。）

# 满分代码 #

#include<iostream>
    #include<vector>
    using namespace std;
    struct node{
        int id,data,next;
    };
    int main(){
        int begin, n, k, s, d, e;
        cin>>begin>>n>>k;
        node a[100010];
        vector<node> v,ans;
        for(int i=0;i<n;i++){
            scanf("%d%d%d\n",&s,&d,&e);
            a[s]={s,d,e};
            
        }
        for(;begin!=-1;begin=a[begin].next){
            v.push_back(a[begin]);
        }
        for(int i=0;i<v.size();i++){
            if(v[i].data<0)ans.push_back(v[i]);
        }
        for(int i=0;i<v.size();i++){
            if(v[i].data>=0 && v[i].data<=k)ans.push_back(v[i]);
        }
        for(int i=0;i<v.size();i++){
            if(v[i].data>k)ans.push_back(v[i]);
        }
        for(int i=0;i<ans.size();i++){
            if(i!=ans.size()-1)printf("%05d %d %05d\n",ans[i].id,ans[i].data,ans[i+1].id);
            else printf("%05d %d -1",ans[i].id,ans[i].data);
        }
        return 0;
    }

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