题目1519：合并两个排序的链表

题目1519：合并两个排序的链表

时间限制：1 秒

内存限制：128 兆

特殊判题：否

提交：2069

解决：937

题目描述：

输入两个单调递增的链表，输出两个链表合成后的链表，当然我们需要合成后的链表满足单调不减规则。
(hint: 请务必使用链表。)

输入：

输入可能包含多个测试样例，输入以EOF结束。
对于每个测试案例，输入的第一行为两个整数n和m(0<=n<=1000, 0<=m<=1000)：n代表将要输入的第一个链表的元素的个数,m代表将要输入的第二个链表的元素的个数。
下面一行包括n个数t(1<=t<=1000000)：代表链表一中的元素。接下来一行包含m个元素，s(1<=t<=1000000)。

输出：

对应每个测试案例，
若有结果，输出相应的链表。否则，输出NULL。

样例输入：

5 21 3 5 7 92 40 0

样例输出：

1 2 3 4 5 7 9NULL

import java.io.BufferedReader;import java.io.InputStream;import java.io.InputStreamReader;import java.io.PrintWriter;import java.math.BigInteger;import java.util.StringTokenizer;public class Main {    public static void main(String[] args) {    new Task().solve() ;  }}class Task{InputReader in = new InputReader(System.in) ;PrintWriter out = new PrintWriter(System.out) ;class Node{   int value ;   Node next ;   Node(int value){   this.value = value ;   next = null ;   }}Node buildNode(int n){Node head = null , now = null ;    for(int i = 1 ; i <= n ; i++){    Node node = new Node(in.nextInt()) ;    if(head == null) now = head = node ;    else{    now.next = node ;    now = node ;    }    }return head ; }Node merge(Node first , Node second){Node head = null , now = null ;while(first != null && second != null){Node node = null ;if(first.value <= second.value){node = new Node(first.value) ;first = first.next ;}else{node = new Node(second.value) ;second = second.next ;}if(head == null) now = head = node ;else{now.next = node ;now = now.next ;}}while(first != null){Node node = new Node(first.value) ;first = first.next ;if(head == null) now = head = node ;else{now.next = node ;now = now.next ;}}while(second != null){Node node = new Node(second.value) ;second = second.next ;if(head == null) now = head = node ;else{now.next = node ;now = now.next ;}}return head ; }void printNode(Node head){if(head == null){out.println("NULL") ;return ;}else{Node now = head ;while(now != null){out.print(now == head ? "" : " ") ;out.print(now.value) ;now = now.next ;}out.println();}}void solve(){while(in.hasNext()){int n = in.nextInt() ;int m = in.nextInt() ;Node first = buildNode(n) ;Node second = buildNode(m) ;printNode(merge(first, second)) ;//out.flush();}out.flush();}}class InputReader {      public BufferedReader reader;      public StringTokenizer tokenizer;        public InputReader(InputStream stream) {          reader = new BufferedReader(new InputStreamReader(stream), 32768);          tokenizer = new StringTokenizer("");      }        private void eat(String s) {          tokenizer = new StringTokenizer(s);      }        public String nextLine() {          try {              return reader.readLine();          } catch (Exception e) {              return null;          }      }        public boolean hasNext() {          while (!tokenizer.hasMoreTokens()) {              String s = nextLine();              if (s == null)                  return false;              eat(s);          }          return true;      }        public String next() {          hasNext();          return tokenizer.nextToken();      }        public int nextInt() {          return Integer.parseInt(next());      }        public long nextLong() {          return Long.parseLong(next());      }        public double nextDouble() {          return Double.parseDouble(next());      }        public BigInteger nextBigInteger() {          return new BigInteger(next());      }    }