PTA 5-2 Reversing Linked List (25) [法二] - <algorithm> - reverse()

题目：http://www.patest.cn/contests/pat-a-practise/1074

转自：http://blog.csdn.net/xtzmm1215/article/details/43195793

法一：http://www.cnblogs.com/claremore/p/4802164.html

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         //第一行：链表的首地址add，结点个数n，每隔k个进行一次反转
00000 4 99999     //后面n行：结点的地址address，数据data，下一个结点的地址next
00100 1 12309
68237 6 -1
33218 3 00000
99999 5 68237
12309 2 33218

Sample Output:

00000 4 33218     //反转之后的结果
33218 3 12309
12309 2 00100
00100 1 99999
99999 5 68237
68237 6 -1

题目大意：

反转单链表，给定常数K和单链表L，要求按每K个节点反转单链表，如：L: 1->2->3->4->5->6 K=3,输出:3->2->1->6->5->4，如果K=4,输出：4->3->2->1->5->6.

特殊情况：

1. 有其他链表干扰（即有多个-1出现）：找链表有效长度

2. k=1 或者 k=n ：不变 或者 全部逆序

3. 需逆序的起点大于有效长度 : 直接输出原链表

推荐测试：

1. k=1或者k=n：

00100 6 6
00000 4 99999
00100 1 12309
68237 6 -1
33218 3 00000
99999 5 68237
12309 2 33218

2. 有其他链表干扰（即有多个-1出现）：

00100 6 2
00000 4 99999
00100 1 12309
68237 6 -1
33218 3 -1
99999 5 68237
12309 2 33218

代码：

#include <stdio.h>
#include <vector>
#include <algorithm>
using namespace std;
 
#define MAXN 100001
typedef struct{
    int addr;
    int data;
    int next;
}Node;
Node nodes[MAXN];
vector<Node> list;
 
int main()
{   
    int firstAdd, n, k;  
    scanf("%d%d%d", &firstAdd, &n, &k);
    while(n--){
        Node nn;
        scanf("%d%d%d", &nn.addr, &nn.data, &nn.next);
        nodes[nn.addr] = nn;  //方便存入结构体数组
    }
    int address = firstAdd;
    while(address != -1){
        //按地址依次存入list中
        list.push_back(nodes[address]);
        address = nodes[address].next;
    }
    int length = list.size(); //有效长度（address == -1 结束）
    int round = length/k;     //需逆序几组
    for(int i = 1; i <= round; ++i){  
        int start = (i-1)*k;
        int end = i*k;  
        reverse(list.begin() + start, list.begin() + end); //#include <algorithm>
    }
    for(int i = 0; i < length-1; ++i){   
        printf("%05d %d %05d\n", list[i].addr, list[i].data, list[i+1].addr);
    }
    //最后一个节点的next为-1
    printf("%05d %d %d\n",list[length-1].addr, list[length-1].data, -1); 
    return 0;
}