[LeetCode]23. Merge k Sorted Lists

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

Example:

Input:
[
1->4->5,
1->3->4,
2->6
]
Output: 1->1->2->3->4->4->5->6
要減少算法的複雜度，就要利用已知的條件，已知的條件就是已有的順序。

public ListNode mergeKLists(ListNode[] lists) {
    ListNode trueHead=new ListNode(0);
    if(lists.length<=0) return trueHead.next;
    trueHead.next=lists[0];
    for(int i=1;i<lists.length;i++){
        ListNode cur=lists[i];
        ListNode pre=trueHead;
        while(cur!=null){
            while(pre.next!=null && pre.next.val<=cur.val) pre=pre.next;
            ListNode next=pre.next;
            ListNode curNext=cur.next;
            pre.next=cur;
            cur.next=next;
            cur=curNext;
            pre=pre.next;
        }
    }
    return trueHead.next;
}

咱們一共會作k-1次合併，能夠經過歸併操做減小到logk次

public ListNode mergeKLists(ListNode[] lists) {
        if(lists.length<=0) return null;
        int len=1;
        while(len<=lists.length){
            for(int i=0;i<lists.length-len;i=i+2*len){
                ListNode left=lists[i];
                ListNode cur=lists[i+len];
                ListNode trueHead=new ListNode(0);
                trueHead.next=left;
                ListNode pre=trueHead;
                while(cur!=null){
                    while(pre.next!=null && pre.next.val<=cur.val) pre=pre.next;
                    ListNode next=pre.next;
                    ListNode curNext=cur.next;
                    pre.next=cur;
                    cur.next=next;
                    cur=curNext;
                    pre=pre.next;
                }
                lists[i]=trueHead.next;
            }
            len*=2;
        }
        return lists[0];
    }