1.该题要求求出翻转的链表,与leetcode中的Reverse Linked List 和Reverse Linked List II相似。
2.这次解法并没有采用链表操作,而是把链表转化为数组,对数组进行操作。
Given a constant K and a singly linked list L, you are supposed to reverse the links of every K elements on L. For example, given L being 1→2→3→4→5→6, if K = 3, then you must output 3→2→1→6→5→4; if K = 4, you must output 4→3→2→1→5→6.
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 (<=N) which is the length of the sublist to be reversed. 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, and Next is the position of the next node.
Output Specification:
For each case, output the resulting ordered linked list. Each node occupies a line, and is printed in the same format as in the input.
Sample Input:
00100 6 4 00000 4 99999 00100 1 12309 68237 6 -1 33218 3 00000 99999 5 68237 12309 2 33218
Sample Output:
00000 4 33218 33218 3 12309 12309 2 00100 00100 1 99999 99999 5 68237 68237 6 -1
AC代码:
//#include<string> //#include <iomanip> //#include<stack> //#include<unordered_set> //#include <sstream> //#include "func.h" //#include <list> #include<unordered_map> #include<set> #include<queue> #include<map> #include<vector> #include <algorithm> #include<stdio.h> #include<iostream> #include<string> #include<memory.h> #include<limits.h> #include<stack> using namespace std; /* 00100 6 4 00000 4 99999 00100 1 12309 68237 6 -1 33218 3 00000 99999 5 68237 12309 2 33218 00100 6 3 00000 4 99999 00100 1 12309 68237 6 -1 33218 3 00000 99999 5 68237 12309 2 33218 00100 6 2 00000 4 99999 00100 1 12309 68237 6 -1 33218 3 00000 99999 5 68237 12309 2 33218 00100 6 0 00000 4 99999 00100 1 12309 68237 6 -1 33218 3 00000 99999 5 68237 12309 2 33218 00100 6 1 00000 4 99999 00100 1 12309 68237 6 -1 33218 3 00000 99999 5 68237 12309 2 33218 00100 6 5 00000 4 99999 00100 1 12309 68237 6 -1 33218 3 00000 99999 5 68237 12309 2 33218 00100 6 6 00000 4 99999 00100 1 12309 68237 6 -1 33218 3 00000 99999 5 68237 12309 2 33218 */ struct ListNode{ int val; int add; ListNode*next; ListNode(int x) :val(x), next(NULL){}; ListNode() :val(0), next(NULL){}; }; int main(void) { vector<ListNode> list(100001,ListNode(-1)); int head, recordSum, step; cin >> head >> recordSum >> step; for (int i = 0; i < recordSum; i++) { int now, val, next; scanf("%d %d %d", &now, &val, &next); list[now].val = val; list[now].add = now; if (next != -1) { list[now].next = &list[next]; list[next].add = next; } } vector<int> num(100001, -1);//记录地址,地址是唯一的 int idx=0; ListNode* root = &list[head]; while (root) { num[idx++] = root->add; root = root->next; } if (step!=0)//等于0时,会进入死循环 for (int i = 0; i+step <=idx; i+=step) { for (int j = 0; j < step / 2; j++) { swap(num[j+i], num[step + i - 1 - j]); } } for (int i = 0; i < idx; i++) { if (i != idx - 1) printf("%05d %d %05d\n", num[i], list[num[i]].val, num[i + 1]); else printf("%05d %d -1\n", num[i], list[num[i]].val); } return 0; }