PAT甲级练习1074. Reversing Linked List (25)

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
用int数组保存位置信息,用next信息判断在链中的点,然后根据位置信息按照规则变换即可

#include   
#include   
#include   
#include   
#include   
#include   
#include   
#include   
#include   
#include  
using namespace std;

const int MAX = 1e5+10;
int start, nn[MAX], data[MAX], use[MAX], result[MAX];

int main() {

	int n, k, num=0;
	scanf("%d%d%d", &start, &n, &k);

	for(int i=1; i<=n; i++){
		int l, d, nt;
		scanf("%d%d%d", &l, &d, &nt);
		nn[l] = nt;
		data[l] = d;
	}

	while(start!=-1){
		use[++num] = start, result[num] = use[num], start = nn[start];
	}
	int times=num/k, num1=0;
	for(int i=1; i<=times; i++){
		for(int j=i*k; j>=(i-1)*k+1; j--){
			result[++num1] = use[j]; 
		}
	}
	for(int i=1; i

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