[leetcode]23. Merge k Sorted Lists

Merge k sorted linked lists and return it as one sorted list. Analyze and describe its complexity.

Leetcode 21题是合并两个链表。合并K个链表的方法就是2的进一步。有两种方法：

1 归并法。即每次链表数组中相邻的两个链表合并，结果为一个新的链表数组，重复之前的流程知道数组中只有一个元素为止。

2 按部就班法。从链表数组的第一个链表不断和后面的链表合并存到第一个链表处，最后返回第一个链表。

java代码：方法1

/** * Definition for singly-linked list. * public class ListNode { *     int val; *     ListNode next; *     ListNode(int x) { val = x; } * } */public class Solution {    public ListNode mergeKLists(ListNode[] lists) {       ListNode[] res =  mergek(lists);       if(res.length==0){           return null;       }       else{           return res[0];       }    }    public ListNode[] mergek(ListNode[] list){        if(list.length==0||list.length==1){            return list;        }        else{            int len = list.length;            int newLength = 0;            if(len%2==0){                newLength=len/2;            }            else{                newLength=len/2+1;            }            ListNode[] tmp = new ListNode[newLength];            int index=0;            for(int i=0;i<len;i+=2){                if((i+1)<len){                 tmp[index++]=mergeTwoLists(list[i],list[i+1]);                }                else{                    tmp[index++]=list[i];                }            }            return mergek(tmp);        }    }     public ListNode mergeTwoLists(ListNode l1, ListNode l2) {                if(l1==null&&l2==null)        {            return null;        }        if(l1==null)        {            return l2;        }        if(l2==null)        {            return l1;        }        ListNode cur=null,head=null;        ListNode node1=l1,node2=l2;        while(node1!=null&&node2!=null)        {            if(node1.val<node2.val)            {                if(cur==null)                {                    cur=node1;                }                else                {                    cur.next=node1;                    cur=node1;                }                node1=node1.next;            }            else            {                if(cur==null)                {                    cur=node2;                }                else                {                    cur.next=node2;                    cur=node2;                }                node2=node2.next;            }                    }        while(node1!=null)        {            cur.next=node1;            cur=node1;            node1=node1.next;        }        while(node2!=null)        {            cur.next=node2;            cur=node2;            node2=node2.next;        }        if(l1.val<l2.val)        {            return l1;        }        else        {            return l2;        }            }}

go代码：方法2

/** * Definition for singly-linked list. * type ListNode struct { *     Val int *     Next *ListNode * } */func mergeKLists(lists []*ListNode) *ListNode {    if len(lists)==0{        return nil    }else if len(lists)==1{        return lists[0]    }else{        for i:=1;i<len(lists);i++{            lists[0]=mergeTwoLists(lists[0],lists[i])        }        return lists[0]    }}func mergeTwoLists(l1 *ListNode, l2 *ListNode) *ListNode {    if l1 == nil && l2 == nil{        return nil    }    if l1==nil{        return l2    }    if l2 == nil{        return l1;    }    var head,cur *ListNode    for l1!=nil&&l2!=nil{        if l1.Val<l2.Val{            if head == nil{                head = l1                cur = l1            }else{                cur.Next = l1                cur=cur.Next            }            l1=l1.Next        }else{            if head == nil{                head = l2                cur = l2            }else{                cur.Next = l2                cur=cur.Next            }            l2=l2.Next        }    }    if l1!=nil{        cur.Next=l1    }    if l2!=nil{        cur.Next=l2    }    return head}