PAT 1074. Reversing Linked List (25)（链表反转）

官网

1074. Reversing Linked List (25)

时间限制
400 ms
内存限制
65536 kB
代码长度限制
16000 B
判题程序
Standard
作者
CHEN, Yue
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

解题思路

• 1.是每k次反转一下，不是前k次反转就行了，我看错了大半年别学我。
• 2.注意k=0，1，链表的长度（不一定为n），以及k>链表的长度这几种情况就可以了。

AC代码

#include<string>#include<iostream>#include<map>#include<cstdio>#include<vector>#include<deque>#include<algorithm>#include<math.h>using namespace std;//00100 6 6//00000 4 99999//00100 1 12309//68237 6 -1//33218 3 00000//99999 5 68237//12309 2 33218int value[100000],nx[100000];int main(int argc, char *argv[]){    int firstAddre,n,k;    cin >> firstAddre >> n >> k;    int tem;    for (int i = 0; i < n; ++i) {        cin >> tem;        cin >> value[tem]>>nx[tem];    }    int now = firstAddre;    //储存    vector<int> keep;    while (now!=-1) {        keep.push_back(now);        now = nx[now];    }    int bot,up ;    int N = 0;    if (k!=0) {        N = keep.size()/k;    }    //每k次反转    for (int j = 0; j < N; ++j) {            for (int i = 1; i <= k/2; ++i) {                up = i-1+k*j;                bot = k-i+k*j;                tem = keep[up];                keep[up] = keep[bot];                keep[bot] = tem;            }    }    //输出    for (int i = 0; i < (int)keep.size()-1; ++i) {        tem = keep[i];        printf("%05d %d %05d\n",tem,value[tem],keep[i+1]);    }    printf("%05d %d -1\n",keep.back(),value[keep.back()]);    return 0;}