PAT 甲级 1133 Splitting A Linked List (25分)

博主 950 2020-07-04

Description:
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

思路：
思路很简单，遍历比较与K和0的关系即可。一开始用vector存储的是node，这样作insert操作消耗时间很大，优化为存储地址值(下标)。

坑点：
但是被PTA的编译器坑了，这题花了很多时间找bug
一开始for循环写的是

for (int i = 0; i < N; ++i) {
        int add;
        cin >> add >> list[add].data >> list[add].next;
    }

在自己的ide运行没毛病，换到pat的编译器(g++6.5.0)就。。。
image.png

想了很久，把for循环改了就Accepted了

for (int i = 0; i < N; ++i) {
        int add;
        cin >> add;
        cin >> list[add].data >> list[add].next;
    }

个人理解是pat的编译器对一个输入流一次性读入数据并赋值，这样add的值读入之后并不是list[add]中的add。
查了一下我使用的xcode的c编译器是使用的clang，把PTA的编译器换成clang++6.0.1
image.png
也可以了。。-_-||

code:

#include<iostream>
#include<unordered_map>
#include<vector>
#include<algorithm>
using namespace std;
const int maxn = 100010;
struct node{
    int data;
    int next;
}list[maxn];


int main(){
    int head, N, K;
    cin >> head >> N >> K;
    for (int i = 0; i < N; ++i) {
        int add;
        cin >> add >> list[add].data >> list[add].next;
    }
    
    //从头结点开始 分别记录 小于0的部分 [0,K]的部分 大于K的部分 的地址(比记录node消耗空间和时间都少)
    vector<int> negativePart;
    vector<int> smallPart;
    vector<int> bigPart;
    
    while (head != -1) {
        if (list[head].data < 0)
            negativePart.push_back(head);
        else if(list[head].data <= K)
            smallPart.push_back(head);
        else
            bigPart.push_back(head);
        head = list[head].next;
    }
    
    //把后两个部分的地址都插入第一个部分后面
    negativePart.insert(negativePart.end(), smallPart.begin(), smallPart.end());
    negativePart.insert(negativePart.end(), bigPart.begin(), bigPart.end());
    for (int i = 0; i < negativePart.size(); ++i) {
        if (i != negativePart.size() - 1) {
            printf("%05d %d %05d\n", negativePart[i], list[negativePart[i]].data, negativePart[i+1]);
        }else printf("%05d %d -1\n", negativePart[i], list[negativePart[i]].data);
        
    }
    return 0;
}
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