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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 (<=1000). 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

题目大意：

给一个链表和K，遍历链表后将<0的结点先输出，再将0～k区间的结点输出，最后输出>k的结点

分析：

将结点用list[10000]保存，list为node类型，node中保存结点的值value和它的next地址。list的下标就是结点的地址。将<0、0～k、>k三部分的结点地址分别保存在v[0]、v[1]、v[2]中，最后将vector中的值依次输出即可～

原文链接：https://blog.csdn.net/liuchuo/article/details/78037305

题解

#include <bits/stdc++.h>

using namespace std;
const int maxn=100000;
struct Node
{
    int address,key,next;
}node[maxn];
int vis[maxn];
int main()
{
#ifdef ONLINE_JUDGE
#else
    freopen("1.txt", "r", stdin);
#endif
    int n,s,id,k;
    cin>>s>>n>>k;
    for(int i=0; i<n; i++)
    {
        cin>>id;
        node[id].address=id;
        cin>>node[id].key>>node[id].next;
    }
    vector<int> v[3];
    for(int i=s;i!=-1;i=node[i].next){
        if(node[i].key<0){
            v[0].push_back(i);
        }else if(node[i].key>k){
            v[2].push_back(i);
        }else v[1].push_back(i);
    }
    int flag=0;
    for (int i = 0; i < 3; i++) {
        for (int j = 0; j < v[i].size(); j++) {
            if (flag == 0) {
                printf("%05d %d ", v[i][j], node[v[i][j]].key);
                flag = 1;
            } else {
                printf("%05d\n%05d %d ", v[i][j], v[i][j], node[v[i][j]].key);
            }
        }
    }
    printf("-1");
    return 0;
}

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