java链表面试编程题

通过编码实现日常关于链表可能会遇到到编程题

1 两个链表是各自自增的,要求合拼之后的链表满足单调不递减

/**
 * 两个递增的单链表合并保持单调递增
 *
 * 递归求解
 * @param firstNode
 * @param secondNode
 * @return
 */
public Node mergeNode(Node firstNode,Node secondNode){
    if (firstNode==null)return secondNode;
    if (secondNode==null)return firstNode;
    Node mergeNode = null;
    if (firstNode.datadata){
        mergeNode = firstNode;
        mergeNode.next = mergeNode(firstNode.next,secondNode);
    }
    else {
        mergeNode = secondNode;
        mergeNode.next = mergeNode(firstNode,secondNode.next);
    }
    return mergeNode;
}

2 单链表排序

/**
 * 单链表排序
 * 归并排序
 */
public Node sortNode(Node head){
    if(head == null|| head.next == null){
        return head;
    }
    Node mid = middleNode(head);
    Node right = sortNode(mid.next);
    mid.next = null;
    Node left = sortNode(head);
    return merge(left, right);
}

/**
 * 合拼排好序的子链表
 * @param n1
 * @param n2
 * @return
 */
private Node merge(Node n1,Node n2){
    Node dummy = new Node(0);
    Node node = dummy;
    while (n1!=null&&n2!=null) {
        if(n1.datadata){
            node.next = n1;
            n1 = n1.next;
        }else{
            node.next = n2;
            n2 = n2.next;
        }
        node = node.next;
    }
    if(n1!=null){
        node.next = n1;
    }else{
        node.next = n2;
    }
    return dummy.next;
}

/**
 * 寻找链表中间值
 * @param head
 * @return
 */
private Node middleNode(Node head){
    Node slow = head;
    Node fast = head.next;
    while(fast!=null&fast.next!=null){
        slow = slow.next;
        fast = fast.next.next;
    }
    return slow;
}

3 单链表整体反转

/**
 * 递归方式-进行反转
 * @param node
 * @return
 */
private Node reverseNodeByDG(Node head){
    if(head.next == null){
        return head;
    }
    Node reverseNode = reverseNodeByDG(head.next);
    head.next.next = head;
    head.next = null;
    return reverseNode;
}

/**
 * 遍历方式-进行反转
 * @param node
 * @return
 */
private Node reverseNodeByBL(Node node){
    Node prev = null;
    while(node!=null){
        Node tmp = node.next;
        node.next = prev;
        prev = node;
        node = tmp;
    }
    return prev;
}

4 单链表从n到m位置的反转

/**
 * 翻转链表的n到m之间的节点
 */
Node reverseFromNTOM(Node head,int m,int n){
    if(m>=n||head == null){
        return head;
    }
    Node dummy = new Node(0);
    dummy.next = head;
    head = dummy;
    for(int i = 1;i;i++){
        if(head == null){
            return null;
        }
        head = head.next;
    }
    Node pmNode = head;
    Node mNode = head.next;
    Node nNode = mNode;
    Node pnNode = mNode.next;
    for(int i = m;i;i++){
        if(pnNode == null){
            return null;
        }
        Node tmp = pnNode.next;
        pnNode.next = nNode;
        nNode = pnNode;
        pnNode = tmp;
    }
    pmNode.next = nNode;
    mNode.next = pnNode;
    return dummy.next;
}

5 单链表根据指定值进行分割

/**
 * 单链表分区
 * 小于x的结点排在大于或等于x的结点之前
 */
public Node nodePartition(Node head,int x){
    if(head == null){
        return null;
    }
    Node left = new Node(0);
    Node right = new Node(0);

    Node leftNode = left;
    Node rightNode = right;
    while(head!=null){
        if(head.datanext = head;
            left = head;
        }else{
            right.next = head;
            right = head;
        }
        head = head.next;
    }
    left.next = rightNode.next;
    right.next = null;
    return leftNode.next;
}