目录

题目一：合并两个有序链表

        代码：

题目二：合并k个有序链表

        代码：

题目三：判断链表中是否有环

        代码：

题目四：链表中环的入口结点

        代码：

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题目一：合并两个有序链表

        代码：

import java.util.*;/** public class ListNode {*   int val;*   ListNode next = null;*   public ListNode(int val) {*     this.val = val;*   }* }*/public class Solution {/*** 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可** * @param pHead1 ListNode类 * @param pHead2 ListNode类 * @return ListNode类*/public ListNode Merge (ListNode pHead1, ListNode pHead2) {if(pHead1==null) return pHead2;if(pHead2==null) return pHead1;// write code hereListNode result=null;ListNode index1=pHead1;ListNode index2=pHead2;if(index1.val<index2.val){result=index1;index1=index1.next;}else{result=index2;index2=index2.next;}ListNode last=result;while(index1!=null && index2!=null){if(index1.val<index2.val){last.next=index1;index1=index1.next;}else{last.next=index2;index2=index2.next;}last=last.next;}if(index1==null){last.next=index2;}if(index2==null){last.next=index1;}return result;}
}

题目二：合并k个有序链表

        代码：

        利用的是“归并排序”的思路。

import java.util.*;/** public class ListNode {*   int val;*   ListNode next = null;*   public ListNode(int val) {*     this.val = val;*   }* }*/public class Solution {/*** 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可*** @param lists ListNode类ArrayList* @return ListNode类*/public ListNode mergeKLists (ArrayList<ListNode> lists) {// write code herereturn divideMerge(lists,0,lists.size()-1);}//划分合并区间public ListNode divideMerge(ArrayList<ListNode> lists,int left,int right){if(left>right) return null;else if(left==right) return lists.get(left);else{int mid=(left+right)/2;return Merge(divideMerge(lists,left,mid),divideMerge(lists,mid+1,right));}}//合并两个有序链表public ListNode Merge (ListNode pHead1, ListNode pHead2) {if (pHead1 == null) return pHead2;if (pHead2 == null) return pHead1;// write code hereListNode result = null;ListNode index1 = pHead1;ListNode index2 = pHead2;if (index1.val < index2.val) {result = index1;index1 = index1.next;} else {result = index2;index2 = index2.next;}ListNode last = result;while (index1 != null && index2 != null) {if (index1.val < index2.val) {last.next = index1;index1 = index1.next;} else {last.next = index2;index2 = index2.next;}last = last.next;}if (index1 == null) {last.next = index2;}if (index2 == null) {last.next = index1;}return result;}
}

题目三：判断链表中是否有环

        代码：

        利用的是双指针（快慢指针）的思路。

import java.util.*;
/*** Definition for singly-linked list.* class ListNode {*     int val;*     ListNode next;*     ListNode(int x) {*         val = x;*         next = null;*     }* }*/
public class Solution {public boolean hasCycle(ListNode head) {if(head==null || head.next==null){return false;}ListNode fast=head;ListNode low=head;while(fast!=null && fast.next!=null){fast=fast.next.next;low=low.next;if(fast==null) return false;if(fast==low) return true;}return false;}
}

题目四：链表中环的入口结点

        代码：

        快慢指针的思路

import java.util.*;
/*public class ListNode {int val;ListNode next = null;ListNode(int val) {this.val = val;}
}
*/
public class Solution {public ListNode EntryNodeOfLoop(ListNode pHead) {if (pHead == null || pHead.next == null) {return null;}ListNode fast = pHead;ListNode low = pHead;while (fast != null && fast.next != null) {fast = fast.next.next;low = low.next;if (fast == low) break;}if(fast==null) return null;low=pHead;while(low != fast){fast=fast.next;low=low.next;}return fast;}
}