单链表的初阶练习

.连接两个升序链表
1.建立一个节点负责串起新的表；
2.当两个头节点都不为空时，判断头两节点数据的大小；
小的被新建节点串起，然后节点后移；
3.当有一条链被串完时，直接串另一个节点；

public static Node mergeTwoLists(Node headA, Node headB) {
    Node newHead = new Node();
    Node tmp = newHead;
    while (headA != null && headB != null){
        if(headA.data < headB.data) {
            newHead.next = headA;
            headA = headA.next;
            tmp = tmp.next;
        }else {
            newHead.next = headB;
            headB = headB.next;
            tmp = tmp.next;
        }
    }
    if(headA!= null) {
        tmp.next = headA;
    }
    if(headB != null) {
        tmp.next = headB;
        }
        return newHead.next;
        }
   

返回两个链表的第一个交点

1.得的出两链表的长度差值len
2.建立两个节点，让其中一个节点先走len步
3.两个节点同时走直到他们相等

public static Node getIntersectionNode(Node headA,Node headB){
    int lenA = 0;
    int lenB = 0;
    Node pL = headA;//长的
    Node pS = headB;//短的
    while (pL != null) {
        lenA++;
        pL = pL.next;
    }
    while (pS != null) {
        lenB++;
        pS = pS.next;
    }
    
    pL = headA;
    pS = headB;
    int len = lenA - lenB;
    if(len < 0) {
        pL = headB;
        pS = headA;
        len = lenB-lenA;
    }

    //pL指向最长的单链表
    //pS指向最短的单链表
    //len  肯定是一个正数
    for (int i = 0; i < len; i++) {
        pL = pL.next;
    }
    //pL和pS此时已经在同一期起跑线上了
    //可以一人一步走

    while (pL != null && pS != null && pL != pS) {
        pL = pL.next;
        pS = pS.next;
    }
    if(pL == null || pS == null ) {
        return null;
    }
    return  pL;
}

链表是否有为回文结果（形似123454321）
1.创建快慢两个指针，分别到达链表的尾部和中间
2.快指针返回头部，慢指针在中间开始对后半段进行反转
3.后移头节点和慢指针并比较其内的数值，有不相等则不是回文结果

public static boolean chkPalindrome(Node head){//是否回文
    if(head == null ){//空表
        return false;
    }
    if(head.next == null) {//单节点
        return true;
    }
    Node fast = head;
    Node slow = head;
    while (fast != null && fast.next != null) {
        fast = fast.next.next;
        slow = slow.next;
    }
    //2、翻转
    Node p = slow.next;
    while (p != null) {
        Node pNext = p.next;
        p.next = slow;
        slow = p;
        p = pNext;
        if(p != null) {
            pNext = p.next;
        }
    }
    //3、进行判断
    while (head != slow) {
        if(head.data != slow.data) {
            return false;
        }
        if(head.next == slow) {
            return true;
        }
        head = head.next;
        slow = slow.next;
    }
    return true;
}