1133 Splitting A Linked List (25分)-优快云博客

本文介绍了一种链表操作算法，该算法将链表中的元素按特定条件进行分类重组，确保所有负数出现在非负数之前，且值在[0,K]范围内的元素位于大于K的元素之前，同时保持各分类内元素的原始顺序不变。

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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

#include<iostream>
#include<vector>
using namespace std;
struct node
{
    int add;
    int data;
    int next;
};


int main(){
    int address, n, k;
    int b[100000]; //保存数据
    int a[100000]; //保存NEXT地址
    int flag [100000] = {0};


    cin >> address >> n >> k;
    vector<node> vec;

    for(int i = 0; i < n; i++){
        int temp_a, data, next;
        scanf("%d %d %d", &temp_a, &data, &next);
        a[temp_a] = next;
        b[temp_a] = data;

    }
    int add = address;
    while (add != -1)
    {
        if(b[add] < 0){
            // printf("%05d %d %05d\n", add, b[add], a[add]);
            if(vec.size())
                vec[vec.size() - 1].next = add;
            node temp;
            temp.add = add;
            temp.data = b[add];
            vec.push_back(temp);
            
            flag[add] = 1;
        }
        add = a[add];
    }
    add = address;
    while (add != -1)
    {
        if(b[add] <= k && flag[add] == 0){
            // printf("%05d %d %05d\n", add, b[add], a[add]);
            if(vec.size())
                vec[vec.size() - 1].next = add;
            node temp;
            temp.add = add;
            temp.data = b[add];
            vec.push_back(temp);
            flag[add] = 1;
        }
        add = a[add];
    }
    add = address;
    while (add != -1)
    {
        if(flag[add] == 0){
            // printf("%05d %d %05d\n", add, b[add], a[add]);
            if(vec.size())
                vec[vec.size() - 1].next = add;
            node temp;
            temp.add = add;
            temp.data = b[add];
            vec.push_back(temp);
            flag[add] = 1;
        }
        add = a[add];
    }
    vec[vec.size() - 1].next = -1;

    for(int i = 0; i < vec.size() - 1; i++){
        printf("%05d %d %05d\n", vec[i].add, vec[i].data, vec[i].next);
    }
    printf("%05d %d -1\n", vec[vec.size() - 1].add, vec[vec.size() - 1].data);
    return 0;
}