Leetcode 234.回文链表 JavaScript (Day 9)

js一刷 自解法一

var isPalindrome = function(head) {
    let p1=head 
    let temp=null;
    while(p1.next){
    p1.prev=temp;
    temp=p1;
    p1=p1.next;
    }
    p1.prev=temp;
    let p2=head
    while(p1!=p2){
        if(p1.val!=p2.val) return false;
        p1=p1.prev;
        p2=p2.next;
    } 
    return true;
};

缺点:修改了原链表

js一刷 法二

//找中间结点
 function FindMiddle(head){
    let slow=head,fast=head;
    while(fast!=null&&fast.next!=null){
        slow=slow.next;
        fast=fast.next.next;
    }
    return slow;
 }
//反转链表
function ReserveList(head){
    let rehead=null,cur=head;
    while(cur){
        let temp=cur.next;
        cur.next=rehead;
        rehead=cur;
        cur=temp;
    }
    return rehead; 
}

var isPalindrome = function (head) {
    let flag=true
    let midNode=FindMiddle(head);
    let p1=ReserveList(midNode);
    let p2=head;
    while(p1){
        if(p1.val!=p2.val){
            flag=false;
            break;
        }
        p1=p1.next;
        p2=p2.next;
    }
    //复原链表
    ReserveList(midNode);
    return flag;
};

算法核心:分为四部分
1.找中间节点(快慢指针)
2.反转链表(头插)
3.双指针变量
4.再次反转复原链表