面试题目——《CC150》链表

 

面试题2.1：编写代码，移除未排序链表中的重复结点　　进阶：如果不得使用临时缓冲区，该怎么解决？

package cc150;

import java.util.HashMap;
import java.util.Map;


public class DeleteDups {

	public static void main(String[] args) {
		// TODO 自动生成的方法存根
		LinkedListNode Node0 = new LinkedListNode(1);
		LinkedListNode Node1 = new LinkedListNode(1);
		LinkedListNode Node2 = new LinkedListNode(2);
		LinkedListNode Node3 = new LinkedListNode(3);
		Node0.next = Node1;
		Node1.next = Node2;
		Node2.next = Node3;
		deleteDups(Node0);
		LinkedListNode temp = Node0;
		while(temp != null){
			System.out.println(temp.iData);
			temp = temp.next;
		}
	}
	
//	public static void deleteDups(LinkedListNode n){
//		Map map = new HashMap();
//		LinkedListNode pre = null;	//记录上一次不重复的结点
//		while(n != null){
//			if(map.containsKey(n.iData)){
//				pre.next = n.next;
//			}else{
//				map.put(n.iData, true);
//				pre = n;
//			}
//			n = n.next;
//		}
//		
//	}
	
	public static void deleteDups(LinkedListNode head){
		if(head == null)
			return;
		LinkedListNode current = head;	
		while(current != null){
			LinkedListNode runner = current;
			while(runner.next != null){	//是runner.next
				if(runner.next.iData == current.iData)	//判断内层所有结点是否等于外层结点
					runner.next = runner.next.next;
				else
					runner = runner.next;
			}
			current = current.next;
		}
	}

}

 

面试题2.2：实现一个算法，找出单向链表中倒数第k个结点。——《剑指Offer》面试题15 (找出) &《Leetcode》removeNthNode （移除）

 

package cc150;

public class NthToLast {

	public static void main(String[] args) {
		// TODO 自动生成的方法存根
		ListNode Node1 = new ListNode(1);
		ListNode Node2 = new ListNode(2);
		ListNode Node3 = new ListNode(3);
		Node1.next = Node2;
		Node2.next = Node3;
		System.out.println(nthToLast(Node1,1).val);
	}
	
	public static ListNode nthToLast(ListNode head,int k){	//返回链表的倒数第k个结点
		if(k <= 0)
			return null;
		ListNode p1 = head;
		ListNode p2 = head;
		
		for(int i=0;i

 

面试题2.3：实现一个算法，删除单向链表中间的某个结点，假定你只能访问该结点。

　　思路：因为是单向链表，所以不知道一个结点的前一个结点，所以当这个结点是最后一个的时候，无解

　　　　　只要把要删除的结点的下一个结点的数据复制到这个结点即可。

import java.util.*;
 
/*
public class ListNode {
    int val;
    ListNode next = null;
 
    ListNode(int val) {
        this.val = val;
    }
}*/
public class Remove {
    public boolean removeNode(ListNode pNode) {
        // write code here
        if(pNode == null || pNode.next == null)
            return false;
        ListNode nextNode = pNode.next;//取得要删除结点的下一个结点
        pNode.val = nextNode.val;
        pNode.next = nextNode.next;
        return true;
    }
}

 

 面试题2.4：编写代码，以给定值x为基准讲链表分割成两部分，所有小于x的结点排在大于或者等于x的结点之前。

package cc150;

public class Partition {

	public static void main(String[] args) {
		// TODO 自动生成的方法存根
		ListNode Node1 = new ListNode(5);
		ListNode Node2 = new ListNode(4);
		ListNode Node3 = new ListNode(3);
		ListNode Node4 = new ListNode(2);
		ListNode Node5 = new ListNode(1);
		Node1.next = Node2;
		Node2.next = Node3;
		Node3.next = Node4;
		Node4.next = Node5;
		ListNode temp = partition(Node1,3);
		
		while(temp != null){
			System.out.println(temp.val);
			temp = temp.next;
		}
	}
	
	 public static ListNode partition(ListNode pHead, int x) {
		 // write code here
		 ListNode beforeStart = null;	//记录链表的头结点
		 ListNode afterStart = null;		//记录链表的头结点
		 ListNode beforeEnd = null;
		 ListNode afterEnd = null;
		 
		 ListNode pNext = pHead;
		 while(pNext != null){
			
			 if(pNext.val < x){
				 if(beforeStart == null){
					 beforeStart = pNext;
					 beforeEnd = pNext;
				 }else{
					 beforeEnd.next = pNext;
					 beforeEnd = beforeEnd.next;
				 }
			 }else{
				 if(afterStart == null){
					 afterStart = pNext;
					 afterEnd = pNext;
				 }else{
					 afterEnd.next = pNext;
					 afterEnd = afterEnd.next;
				 }
			 }
			 pNext = pNext.next;
		 }
		 
		 //切记断掉after最后一个元素的next，不然会形成环
		 if(afterEnd != null)
			 afterEnd.next = null;
		 
		 //如果beforeStart为null，返回afterStart头结点
		 if(beforeStart == null)
			 return afterStart;
		
		 beforeEnd.next = afterStart;
		 return beforeStart;
	 }
	 
	 public static class ListNode {
			int val;
			ListNode next;
			ListNode(int x) { val = x; }
		}

}

 

 面试题2.5：给定两个用链表表示的整数，每个结点包含一个数位。这些数位是反向存放的，也就是个位排在链表首部。编写函数对这两个整数求和，并用链表形式返回结果。

　　进阶：假设这些数位是正向存放的，请再做一遍。

package cc150;

import cc150.Partition.ListNode;

public class Plus {

	public static void main(String[] args) {
		// TODO 自动生成的方法存根
		ListNode Node1 = new ListNode(7);
		ListNode Node2 = new ListNode(2);
		ListNode Node3 = new ListNode(3);
		Node1.next = Node2;
		Node2.next = Node3;

		ListNode Node4 = new ListNode(4);
		ListNode Node5 = new ListNode(5);
		ListNode Node6 = new ListNode(6);
		ListNode Node7 = new ListNode(7);
		Node4.next = Node5;
		Node5.next = Node6;
		Node6.next = Node7;
		
		ListNode temp = plusAB(Node1,Node4);
		System.out.println(temp.val);
		while(temp !=null){
			System.out.print(temp.val);
			temp = temp.next;
		}
	}
	
	public static ListNode plusAB(ListNode a, ListNode b) {
        // write code here
		if(a == null && b == null)
			return null;
		int carry = 0;
		ListNode result_temp = new ListNode(0);
		ListNode result = result_temp;
		while(a != null && b != null){
			int value = a.val + b.val + carry;
			
			if(value>=10){
				result_temp.next = new ListNode(value-10);
				carry = 1;
			}else{
				result_temp.next = new ListNode(value);
				carry = 0;
			}
			result_temp = result_temp.next;
			a = a.next;
			b = b.next;
		}
		if(a != null){
			result_temp.next = new ListNode(a.val+carry);
			result_temp.next.next = a.next;
		}
		else if(b != null){
			result_temp.next = new ListNode(b.val+carry);
			result_temp.next.next = b.next;
		}
		else if(carry == 1){
			result_temp.next = new ListNode(carry);
		}
		
		return result.next;
    }
	
	 public static class ListNode {
			int val;
			ListNode next;
			ListNode(int x) { val = x; }
		}

}

 

面试题2.6：给定一个有环链表，实现一个算法返回环路的开头结点。

package cc150;

import cc150.Plus.ListNode;

public class FindBeginning {

	public static void main(String[] args) {
		// TODO 自动生成的方法存根
		ListNode Node1 = new ListNode(1);
		ListNode Node2 = new ListNode(2);
		ListNode Node3 = new ListNode(3);
		ListNode Node4 = new ListNode(4);
		ListNode Node5 = new ListNode(5);
		ListNode Node6 = new ListNode(6);
		ListNode Node7 = new ListNode(7);
		ListNode Node8 = new ListNode(8);
		ListNode Node9 = new ListNode(9);
		
		Node1.next = Node2;
		Node2.next = Node3;
		Node3.next = Node4;
		Node4.next = Node5;
		Node5.next = Node6;
		Node6.next = Node7;
		Node7.next = Node8;
		Node8.next = Node9;
		Node9.next = Node4;
		
		System.out.println(findBeginning(Node1).val);
	}
	
	public static ListNode findBeginning(ListNode head){
		ListNode slow = head;
		ListNode fast = head;
		//slow移动一步，fast移动两步，链表到头或者碰撞的时候停止
		while(fast != null && fast.next != null){
			slow = slow.next;
			fast = fast.next.next;
			if(slow == fast)
				break;
		}
		//如果是链表到头，没有环路
		if(fast == null || fast.next == null)
			return null;
		//将slow指向链表头部，fast指向碰撞处，两者同时移动，必回在环路起点相遇
		slow = head;
		while(slow != fast){
			slow = slow.next;
			fast = fast.next;
		}
		return fast;
	}
	
	 public static class ListNode {
		int val;
		ListNode next;
		ListNode(int x) { val = x; }
	}

}

 

面试题2.7： 编写一个函数，检查链表是否为回文。

　　CC150第一种解法：反转并比较

import java.util.*;
 
/*
public class ListNode {
    int val;
    ListNode next = null;
 
    ListNode(int val) {
        this.val = val;
    }
}*/
public class Palindrome {
    public boolean isPalindrome(ListNode pHead) {//反转并比较，CC150第一种解法
        // write code here
        if(pHead == null || pHead.next == null)
            return true;
        ListNode slow,fast;     //快慢指针法查找链表的中心
        slow = fast = pHead;
        while(fast != null && fast.next != null){
            slow = slow.next;
            fast = fast.next.next;
        }
        if(fast != null){    //链表个数奇数个
            slow.next = reverseList(slow.next);
            slow = slow.next;   //去掉中间的数
        }else{
            slow = reverseList(slow);
        }
        while(slow != null){
            if(pHead.val != slow.val)
                return false;
            slow = slow.next;
            pHead = pHead.next;
        }
        return true;
    }
     
    public ListNode reverseList(ListNode head) {
        if(head ==null){
            return null;
        }
        Stack stack = new Stack();
        ListNode current = head;
        while(current != null){
            stack.add(current);
            current = current.next;
        }
        head = stack.pop();
        current = head;
        while(stack.empty() != true){
            current.next = stack.pop();
            current = current.next;
        }
        current.next = null;  //Memory Limit Exceeded
        return head;
    }
}