16.合并两个排序的链表 Java

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。

解题思路

两种解法：递归和非递归

参考代码

/*
public class ListNode {
    int val;
    ListNode next = null;

    ListNode(int val) {
        this.val = val;
    }
}*/
//递归解法
public class Solution {
    public ListNode Merge(ListNode list1,ListNode list2) {
        if(list1 == null)
            return list2;
        else if(list2 == null)
            return list1;
        ListNode mergehead = null;
        if(list1.val <= list2.val){
            mergehead = list1;
            mergehead.next = Merge(list1.next,list2);
        }else{
            mergehead = list2;
            mergehead.next = Merge(list1, list2.next);
        }
        return mergehead;
    }
}
//非递归解法
public class Solution {
    public ListNode Merge(ListNode list1,ListNode list2) {
        if(list1 == null)
            return list2;
        else if(list2 == null)
            return list1;
        ListNode mergehead = null;
        if(list1.val <= list2.val){
            mergehead = list1;
            list1 = list1.next;
        }else{
            mergehead = list2;
            list2 = list2.next;
        }
        ListNode cur = mergehead;
        while(list1 != null && list2 != null){
            if(list1.val <= list2.val){
                cur.next = list1;
                list1 = list1.next;
            }else{
                cur.next = list2;
                list2 = list2.next;
            }
            cur = cur.next;
        }
        if(list1 == null)
            cur.next = list2;
        else if(list2 == null)
            cur.next = list1;
        return mergehead;
    }
}

同时有一个和这个题类似的题

题目:合并K个排序链表

题目描述

合并 k 个排序链表，返回合并后的排序链表。请分析和描述算法的复杂度。

示例:

输入: [ 1->4->5, 1->3->4, 2->6 ] 输出: 1->1->2->3->4->4->5->6

思路

想办法转成合并两个排序链表再做

代码

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 * ListNode(int x) { val = x; }  * }  */ class Solution { public ListNode mergeKLists(ListNode[] lists) { if(lists.length==0){ return null; } else if(lists.length==1){ return lists[0]; } else if(lists.length==2){ return mergeTwoList(lists[0],lists[1]); }else{ ListNode[] l1=new ListNode[lists.length/2]; ListNode[] l2=new ListNode[lists.length-lists.length/2]; for(int i=0;i<lists.length;i++){ if(i<lists.length/2) l1[i]=lists[i]; else l2[i-lists.length/2]=lists[i]; } return mergeTwoList(mergeKLists(l1),mergeKLists(l2)); } } public ListNode mergeTwoList(ListNode l1,ListNode l2){ if(l1==null){ return l2; } if(l2==null){ return l1; } if(l1.val>=l2.val){ l2.next=mergeTwoList(l1,l2.next); return l2; }else{ l1.next=mergeTwoList(l1.next,l2); return l1; } } }

转载于:https://www.cnblogs.com/feicheninfo/p/10535260.html