02-线性结构1. Reversing Linked List (25)
2015-03-27 10:15
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题目来源:
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:
Sample Output:
解法1: 有一个测试点通不过,原因是运行超时
解法2:参考http://blog.csdn.net/xtzmm1215/article/details/43195793
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
解法1: 有一个测试点通不过,原因是运行超时
import java.util.HashMap; import java.util.Scanner; public class Main{ static class Node { int value ; String next ; public Node(int value,String next) { this.value=value; this.next=next; } public int getValue() { return value; } public void setValue(int value) { this.value = value; } public String getNext() { return next; } public void setNext(String next) { this.next = next; } } /** *(方法说明) *@param 无 *@return 无 *@throws 无 */ public static void main(String[] args){ Scanner scanner = new Scanner(System.in); //头节点地址 String head =scanner.next(); //总共的节点数 int n = scanner.nextInt(); //转换因子 k int k = scanner.nextInt(); //存储节点 HashMap<String,Node> hashMap = new HashMap<String,Node>(); for(int i=0;i<n;++i) { String key = scanner.next(); int value = scanner.nextInt(); String next = scanner.next(); Node node = new Node(value,next); hashMap.put(key, node); } //存储要输出的4个address String[] address =new String[k]; //address的下标 int index =0; address[index++]=head; Node node = hashMap.get(head); //下一个地址的指针 String next = node.next; //第一次输出 boolean firsline=false; while(!next.equals("-1")) { if(index==k) { for(int i=(k-1);i>=0;--i) { if(i!=k-1 || firsline) System.out.println(address[i]); System.out.print(address[i]+" "+hashMap.get(address[i]).value+" "); firsline = true; } index=0; } node = hashMap.get(next); address[index++]= next; next = node.next; } if(index>0) { if(index==k) { for(int i=(k-1);i>=0;--i) { if(i!=k-1 || firsline) System.out.println(address[i]); System.out.print(address[i]+" "+hashMap.get(address[i]).value+" "); firsline = true; } index=0; } else { for(int i=0;i<index;++i) { System.out.println(address[i]); System.out.print(address[i]+" "+hashMap.get(address[i]).value+" "); } } } System.out.println("-1"); } }
解法2:参考http://blog.csdn.net/xtzmm1215/article/details/43195793
#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; } int address = firstAdd; while(address != -1){ list.push_back(nodes[address]); address = nodes[address].next; } int length = list.size(); int round = length/k; for(int i = 1; i <= round; ++i){ int start = (i-1)*k; int end = i*k; reverse(list.begin() + start, list.begin() + end); } for(int i = 0; i < length-1; ++i){ printf("%05d %d %05d\n", list[i].addr, list[i].data, list[i+1].addr); } printf("%05d %d %d\n",list[length-1].addr, list[length-1].data, -1); return 0; }
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