关于单链表的习题练习（牛客网）（力扣）

//点击标题可以到达题目的地址。

1、给定一个链表，返回链表开始入环的第一个节点。 如果链表无环，则返回 null

public class LinkedList {
    public LinkedNode head;
    public LinkedNode detectCycle() {
        LinkedNode fast = this.head;
        LinkedNode slow = this.head;
        while (fast != null && fast.next != null) {
            fast = fast.next.next;
            slow = slow.next;
            if (fast == slow) {
                break;
            }
        }
            if (fast == null || fast.next == null) {
                return null;
            }
            fast = this.head;
            while (fast != slow) {
                fast = fast.next;
                slow = slow.next;
            }
            return slow;
    }
}

2、给定一个链表，判断链表中是否有环 

public class LinkedList {
    public LinkedNode head;
    public boolean hasCycle() {
        if (this.head == null) {
            return false;
        }
        LinkedNode fast = this.head;
        LinkedNode slow = this.head;
        while (fast != null && fast.next != null) {
            fast = fast.next.next;
            slow = slow.next;
            if (fast == slow) {
                return true;
            }
        }
        return false;
    }
}

3、输入两个链表，找出它们的第一个公共结点 

public class Test{//这里的公共节点指的是地址
    public static LinkedNode getIntersectionNode(LinkedNode headA, LinkedNode headB) {
        LinkedNode pl = headA;
        LinkedNode ps = headB;
        int lenA = 0;
        while (pl != null) {
            lenA ++;
            pl = pl.next;
        }
        int lenB = 0;
        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;
        }
        while (len != 0) {
            pl = pl.next;
            len--;
        }
        while (pl != null && ps != null && pl != ps ) {
            pl = pl.next;
            ps = ps.next;
        }
        if (pl != null && pl == ps) {
            return pl;
        }
        return null;
    }

}

 4、链表的回文结构

public class LinkedList {
    public LinkedNode head;
    public boolean chkPalindrome() {
    if (this.head == null) {
        return true;
    }
    LinkedNode fast = this.head;
    LinkedNode slow = this.head;
    while (fast != null && fast.next != null) {
        fast = fast.next.next;
        slow = slow.next;
    }
    LinkedNode cur = slow.next;
    while (cur != null) {
        LinkedNode curNext = cur.next;
        cur.next = slow;
        slow = cur;
        cur = curNext;
    }
    while (this.head != slow) {
        if (this.head.val != slow.val) {
            return false;
        }
        if (this.head.next == slow) {
            return true;
        }
        this.head = this.head.next;
        slow = slow.next;
    }
    return true;
  }
}

 

5、在一个排序的链表中，存在重复的结点，请删除该链表中重复的结点，重复的结点不保留，返回链表头指针

public class LinkedList {
    public LinkedNode head;
    public LinkedNode deleteDuplication() {
        LinkedNode newHead = new LinkedNode(-1);
        LinkedNode tmp = newHead;
        LinkedNode cur = this.head;
        while(cur != null) {
            if (cur.next != null && cur.val == cur.next.val) {
                while (cur.next != null && cur.val == cur.next.val) {
                    cur = cur.next;
                }
                cur = cur.next;
            } else {
                tmp.next = cur;
                tmp = tmp.next;
                cur = cur.next;
            }
        }
        tmp.next = null;
        return newHead.next;
    }
}

 

 

标签：slow,cur,fast,next,力扣,牛客,LinkedNode,习题,null
来源： https://blog.csdn.net/yszhaonan/article/details/120960705