Leetcode 23. 合并 K 个升序链表 (Day 12)

js一刷最佳方法

javascript 复制代码

var merge = function(list1, list2) {
    const dummy = new ListNode(); 
    let cur = dummy; 
    while (list1 && list2) {
        if (list1.val < list2.val) {
            cur.next = list1; 
            list1 = list1.next;
        } else { 
            cur.next = list2; 
            list2 = list2.next;
        }
        cur = cur.next;
    }
    cur.next = list1 ? list1 : list2; 
    return dummy.next;
};
var mergeKLists = function(lists) {
    let len =lists.length;
    if(len===0) return null;
    if(len===1) return lists[0];
    let left=lists.slice(0,Math.floor(len/2));
    let right=lists.slice(Math.floor(len/2),len);
    return merge(mergeKLists(left),mergeKLists(right));
};

核心还是归并排序
时间复杂度 = O(L log m) ,其中L是结点树,m是数组长度 (如图)

js一刷暴力

javascript 复制代码

var merge=function(l1,l2){
    let dummy=new ListNode(-Infinity);
    let cur=dummy;
    while(l1&&l2){
        if(l1.val<l2.val){
            cur.next=l1;
            l1=l1.next;
        }
        else{
            cur.next=l2;
            l2=l2.next;
        }
        cur=cur.next;
    }
    cur.next=l1||l2;
    return dummy.next;
}
var mergeKLists = function(lists) {
    let head=null;
    for(let i=0;i<lists.length;i++){
        head=merge(head,lists[i]);
    }
    return head;
};

时间复杂度 = O(L*m)

js 一刷最小堆

javascript 复制代码

var mergeKLists = function(lists) {
   const heap=new MinPriorityQueue(node=>node.val);
   for(const x of lists){
        if(x) heap.enqueue(x);
   }
   let dummy=new ListNode();
    let cur=dummy;
    while(!heap.isEmpty()){
        let node=heap.dequeue();    
        cur.next=node;
        if(node.next) heap.enqueue(node.next);
        cur=cur.next;
    }
    return dummy.next;
};

要知道最小堆的api MinPriorityQueue(node=>node.val);
下图pq是最小堆

时间复杂度=O(L log k) ,每次入堆和出堆操作的复杂度是log k,总共要进行L次操作,L是总结点数