C/C++面试(4)——链表操作

比较常见面试题目：

1.不带头结点的链表逆序

2.给任意一个结点，在不遍历链表的情况下，删除该节点

参考链接：http://blog.youkuaiyun.com/csufuyi/article/details/3199717

#include<iostream>
#include<stdlib.h>
#include<time.h>
using namespace std;

typedef struct node{
	int value;
	struct node* next;
}Node;

Node* Reverse(Node* head)//但是当时机试题给的接口是void Reverse(Node* head)，这个我就真心想不出来，因为head=p2，程序就死了。
{
/*非递归解法
	if (head == NULL || head -> next == NULL)
		return head;
	Node * p1 = head, *p2 = head -> next, *p3 = NULL;
	p1 -> next = NULL;
	while ((p3 = p2 -> next) != NULL){
		p2 -> next = p1;
		p1 = p2;
		p2 = p3;
	}
	p2 -> next = p1;
	head = p2;
    
	return head;
*/
	//递归
	if (head == NULL || head -> next == NULL)
		return head;
	Node * tail = head -> next;
	Node * newhead = Reverse(head -> next);
	tail -> next = head;
	head -> next = NULL;

	return newhead;
}

void display(Node* head)
{
	Node * cur = head;
	while (cur){
		cout << cur -> value << ' ';
		cur = cur -> next;
	}

	cout << endl;
}

void deleteNode(Node * p)
{
	Node * nextP = p -> next;
	p -> value = nextP -> value;  //复制下一个结点
	p -> next = nextP -> next;
	delete nextP;  //删除下一个结点
}


int main()
{
	int n;
	cin >> n;
	Node * head = NULL, *cur = NULL, *pre = NULL;
	for (int i = 1; i <= n; i ++){
		if (head == NULL){
			head = new Node;
			head -> value = i;
			head -> next = NULL;
			pre = head;
			//cur = pre -> next;
		}
		else{
			cur = new Node;
			cur -> value = i;
			cur -> next = NULL;
			pre -> next = cur;
			pre = cur;
			//cur = cur -> next;
		}
		//cur = pre -> next;
	}
	cur -> next = NULL;

	head = Reverse(head);

	cout << "Reverse:";
	display(head);

	cur = head;
	srand(time(NULL));
	int k = rand() % n;
	for (i = 1; i <= k && cur; i ++)
		cur = cur -> next;

	deleteNode(cur);
	cout << "delete NO." << k << " node :\n";
	display(head);

	return 0;
}