某公司供應鏈部門面試筆試題(需要手寫代碼)

國美供應鏈部門面試筆試題兩道

1： 判斷鏈表是否有環

public class AppTest {

    public static void main(String[] args) {

    }


    public boolean hasCircle(ListNode header) {

        ListNode p1 = header;
        ListNode p2 = header;

        while (true) {
            //  無環p2先到達終點
            if (p2 == null || p2.next == null) {
                return false;
            }

            p2 = p2.next.next;
            p1 = p1.next;

            //  有環p2追上p1了
            if (p1 == p2) {
                return true;
            }
        }
    }

}

class ListNode {

    public ListNode next;

    public ListNode() {
    }

}

需要說明算法複雜度，即要證明p2和p1 一定會在某個節點相遇且相遇時p1走了多少步。

 

2：使用非遞歸實現二叉樹的前序遍歷

public class Main {

    public static void main(String[] args) {

        Node[] result = null ;

        Node node = new Node(1);
        node.left = new Node(2);
        node.right = new Node(3);
        node.left.left = new Node(4);
        node.left.right = new Node(5);
        node.right.left = new Node(6);
        node.right.right = new Node(7);
        result = pre(node) ;
        print(result);

    }


    public static Node[] pre(Node root) {

        if (root == null) {
            return new Node[]{};
        }

        List<Node> nodeList = new ArrayList<>() ;
        Stack<Node> nodeStack = new Stack<>() ;
        nodeStack.push(root) ;

        while (!nodeStack.empty()) {

            Node curItem = nodeStack.peek();

            // 沒有左子樹或者左子樹已經被訪問了  則當前節點出棧
            if (curItem.left == null || curItem.visited ==1){
                nodeStack.pop();
                curItem.visited = 1;
                nodeList.add(curItem);


                Node right = curItem.right ;
                if(right != null){
                    // 右孩子不爲空
                    nodeStack.push(right);
                }else {
                    // 右孩子爲空 則需要標記父節點的左子樹已經被訪問
                    if (!nodeStack.empty()) {
                        curItem = nodeStack.peek();
                        curItem.visited = 1;
                    }
                }
            }

            // 左子樹沒有被訪問  如果左孩子不爲空則壓棧
            while (curItem.visited == 0) {
                Node item = curItem.left ;
                if(item != null) {
                    nodeStack.push(item);
                    curItem = item ;
                }else {
                    break;
                }
            }
        }

        int size = nodeList.size();
        Node[] result = new Node[size];
        nodeList.toArray(result);
        return result;
    }

    public static void print(Node[] result){
        for(Node node:result){
            System.out.println(node.val);
        }
    }
}


class Node {

    public int visited = 0 ;     // 標識左子樹是否被訪問

    public int val ;

    public Node(int val) {
        this.val = val;
    }

    public int getVal() {
        return val;
    }

    public Node left;
    public Node right;
}