链表

23. 合并 K 个升序链表

给你一个链表数组，每个链表都已经按升序排列。

请你将所有链表合并到一个升序链表中，返回合并后的链表。

/**
 * Definition for singly-linked list.
 * function ListNode(val, next) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.next = (next===undefined ? null : next)
 * }
 */
/**
 * @param {ListNode[]} lists
 * @return {ListNode}
 */
var mergeKLists = function(lists) {
    if(!lists||lists.length===0){
        return null
    }
    const mergelist = (app,number) =>{
        if(number === lists.length){
            return app
        }
        let nextnode = lists[number]
        let topnode = app
        let fristnode = app.next
        let secondenode = nextnode
        while(fristnode&&secondenode){
            if(fristnode.val>secondenode.val){
                let sss = secondenode.next
                secondenode.next = topnode.next
                topnode.next = secondenode
                topnode = secondenode
                secondenode = sss
            }else{
                fristnode = fristnode.next
                topnode = topnode.next
            }
        }
        if(fristnode===null){
            topnode.next = secondenode
        }
        return mergelist(app,number+1)
    }
    let mergelistnode = lists[0] || null
    let prelistnode = new ListNode(0,mergelistnode)
    return mergelist(prelistnode,1).next
};

19. 删除链表的倒数第 N 个结点

给你一个链表，删除链表的倒数第 n 个结点，并且返回链表的头结点。

进阶：你能尝试使用一趟扫描实现吗？

/**
 * Definition for singly-linked list.
 * function ListNode(val, next) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.next = (next===undefined ? null : next)
 * }
 */
/**
 * @param {ListNode} head
 * @param {number} n
 * @return {ListNode}
 */
var removeNthFromEnd = function(head, n) {
    let app = new ListNode(0,head)
    let left = app
    let right = app
    while(n--){
        right = right.next
    }
    while(right.next){
        right = right.next
        left = left.next
    }
    left.next = left.next.next
    return app.next
};