【leetcode hot 100 23】合并K个有序链表

解法一：按顺序两两合并升序列表。先排序lists[0]和lists[0]得到结果result，再将结果result与list[3]排序，以此类推。

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public ListNode mergeKLists(ListNode[] lists) {
        if(lists.length==0){
            return null;
        }
        // 按顺序两两合并升序列表
        ListNode result = lists[0];
        for(int i=1;i<lists.length;i++){
            result = mergeTwoLists(result, lists[i]);
        }
        return result;
    }

    public ListNode mergeTwoLists(ListNode list1, ListNode list2) {
        ListNode new_list=null;
        if(list1 != null && list2 != null){
            if(list1.val<=list2.val){
                new_list=list1;
                list1=list1.next;
                new_list.next=null;
            }
            else{
                new_list=list2;
                list2=list2.next;
                new_list.next=null;
            }
        }
        else{
            return list1!=null?list1:list2;
        }
        ListNode curr=new_list;
        while(list1 != null && list2 != null){
            if(list1.val<=list2.val){
                ListNode temp = list1.next;
                curr.next=list1;
                list1.next=null;
                list1=temp;
            }
            else{
                ListNode temp = list2.next;
                curr.next=list2;
                list2.next=null;
                list2=temp;
            }
            curr=curr.next;
        }
        if(list1!=null){
            curr.next=list1;
        }
        if(list2!=null){
            curr.next=list2;
        }
        return new_list;
    }
}

注意：

• 要先判断lists.length==0