PTAMergingLinkedLists

题目

Given two singly linked lists L1​&＃61;a1​→a2​→⋯→an−1​→anL_1​&＃61;a_1​→a_2​→⋯→a_{n−1}​→a_nL1​​&＃61;a1​​→a2​​→⋯→an−1​​→an​​ and L2​&＃61;b1​→b2​→⋯→bm−1​→bmL_2​&＃61;b_1​→b_2​→⋯→b_{m−1}​→b_mL2​​&＃61;b1​​→b2​​→⋯→bm−1​​→bm​​. If n≥2m, you are supposed to reverse and merge the shorter one into the longer one to obtain a list like a1​→a2​→bm​→a3​→a4​→bm−1​⋯a_1​→a_2​→b_m​→a_3​→a_4​→b_m−1​⋯a1​​→a2​​→bm​​→a3​​→a4​​→bm​−1​⋯. For example, given one list being 6→7 and the other one 1→2→3→4→5, you must output 1→2→7→3→4→6→5.

大意

给定两条链表以及他们的首地址&＃xff0c;保证长链表的长度比短链表的两倍和还要长&＃xff0c;将短链表逐个按照每两个元素插入长链表中&＃xff0c;最终输出整个链表。

输入

Each input file contains one test case. For each case, the first line contains the two addresses of the first nodes of L1​L_1​L1​​ and L2L_2L2​​, plus a positive N (≤10510^5105) which is the total number of nodes given. 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 a positive integer no more than 10510^5105, and Next is the position of the next node. It is guaranteed that no list is empty, and the longer list is at least twice as long as the shorter one.

输出

For each case, output in order the resulting linked list. Each node occupies a line, and is printed in the same format as in the input.

输入样例

00100 01000 7
02233 2 34891
00100 6 00001
34891 3 10086
01000 1 02233
00033 5 -1
10086 4 00033
00001 7 -1


输出样例

01000 1 02233
02233 2 00001
00001 7 34891
34891 3 10086
10086 4 00100
00100 6 00033
00033 5 -1


题目分析

首先架构一个结构体数组&＃xff0c;包含该结构的地址&＃xff0c;数据&＃xff0c;下一个地址。采用数组的形式&＃xff0c;数组下标与该结构地址相同&＃xff0c;随后按照地址寻找各数据点下标&＃xff0c;由下标寻觅下一个结点。在输出时注意将上下结点的地址进行修改。

代码

#include
#include
#include
#include
using namespace std;struct Node{int former,data,next;
};
vector<Node> list1,list2;int main(){int head1,head2,n;cin >> head1>>head2>>n;struct Node a[100005];int mark;for(int i &＃61; 0;i<n;&＃43;&＃43;i){cin >> mark;a[mark].former&＃61;mark;//利用下标与代表地址cin>>a[mark].data>>a[mark].next;}int t &＃61; head1;//寻找链表while(t!&＃61;-1){list1.push_back(a[t]);t &＃61; a[t].next;}t &＃61; head2;while(t!&＃61;-1){list2.push_back(a[t]);t &＃61; a[t].next;}if(list1.size()<list2.size()) swap(list1,list2);//确保链表1是长链表&＃xff0c;便于输出t &＃61; list2.size();printf("%05d %d ",list1[0].former,list1[0].data);for(int i &＃61; 1;i<list1.size();&＃43;&＃43;i){printf("%05d\n%05d %d ",list1[i].former,list1[i].former,list1[i].data);if(i%2&＃61;&＃61;1&&t!&＃61;0){//隔2插入printf("%05d\n%05d %d ",list2[t-1].former,list2[t-1].former,list2[t-1].data);--t;}}printf("-1\n");return 0;
}


运行结果