链表 第2天

23.合并k个有序链表

前置知识:合并两个有序链表

常规思路：使用一个变量作为中间变量

    if(lists.length==0) return null
    let res=lists[0]
    for(var i=1;i<lists.length;i++){
        res=mergeKList(res,lists[i])
    }
    return res
};
var mergeKList=function(l1,l2){
let res = new ListNode()
let l3=res
while (l1 && l2) {
   if (l1.val < l2.val) {
       l3.next=l1
       l1 = l1.next
   } else {
       l3.next=l2
       l2 = l2.next
   }
   l3=l3.next
}
l3.next=(l1?l1:l2);
return res.next
}

分治思想

   let n=lists.length
   if(n==0) return null
   if(n==1) return lists[0]
   if (n==2) return mergeKList(lists[0],lists[1])
   let mid =parseInt(n/2)
   let l1=[],
       l2=[]
   for(var i=0;i<mid;i++){
       l1[i]=lists[i]
   }
   for(var i=0;i<n-mid;i++){
       l2[i]=lists[i+mid]
   }
   return mergeKList(mergeKLists(l1),mergeKLists(l2))
};
var mergeKList=function(l1,l2){
    let res = new ListNode()
    let l3=res
    while (l1 && l2) {
        if (l1.val < l2.val) {
            l3.next=l1
            l1 = l1.next
        } else {
            l3.next=l2
            l2 = l2.next
        }
        l3=l3.next
    }
    l3.next=(l1?l1:l2);
    return res.next
}

82.删除链表中的重复元素

var deleteDuplicates = function(head) {
    if(!head) return null
    let res=new ListNode()
    let count=res
    res.next=head
    while(count.next&&count.next.next){
       if(count.next.val==count.next.next.val){
          let x=count.next.val
          while(count.next&&count.next.val==x){
              count.next=count.next.next
          }
       }else{
           count=count.next
       }
    }
    return res.next
};