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：

---

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 here
        ListNode 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 here
        return 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 here
        ListNode 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;
    }
}