鏈表摺疊 面試題

鏈表面試題

我也不知道該叫什麼.就叫 鏈表摺疊 吧

題目:
給一個單鏈表,形如
1 2 3 4 5
變成
1 5 2 4 3

算法複雜度要求爲O(n)

思路:

1. 先把後半部分反轉過來,
2. 然後把後半部分 和 前半部分 合併.

用到的知識點:

1. 構建單鏈表
2. 鏈表反轉
3. 鏈表合併(這個其實相當於插入了一個節點)
4. 遍歷單鏈表

# include <bits/stdc++.h>

using namespace std;

struct node{
	int x;
	struct node * next;
	node(int a,struct node * b):x(a),next(b) {
	}
};

node * head;

//輸入單鏈表 
void input(int n)
{
	int x;
	node * cur;
	for( int i = 0;i < n;i++)
	{
		scanf("%d",&x);
		node *tem = new node(x,NULL);
		if(head == NULL)
		{
			head = tem;
			cur = head;
		}else{
			cur->next = tem;
			cur = cur->next;
		}
		
	}
}

//遍歷單鏈表 
void show(node * p)
{
	printf("\n=============================\n");
	while(p != NULL)
	{
		printf("%d ",p->x);
		p = p->next;
	}
	printf("\n");
}

// 整體反轉 
void reverse(node * p)
{
	node * cur = NULL;
	node * tem ;
	
	while(p != NULL)
	{
		tem = p->next;
		p->next = cur;
		cur = p;
		p = tem;
	}
	head = cur;
}

//鏈表摺疊 
void halfReverse(node * phead) 
{
	node * p = phead;
	
	//鏈表的長度
	int len = 0;
	while(p != NULL)
	{
		len++;
		p = p-> next;
	}
	if(len == 1) return;
	
	int mid = len / 2;
	
	p = phead;
	
	int cnt = 0;
	//找到鏈表一半的位置
	while(p != NULL){
		cnt++;
		if(cnt == mid) break;
		p = p->next;
	}
	
	node * halfHead = p;
	//反轉完成後,cur存放後半部分鏈表的起點
	node * cur = NULL;
	node * tem;
	
	p = p->next;
	while(p != NULL){
		tem = p->next;
		p->next = cur;
		cur = p;
		p = tem;
	}
	
	halfHead->next = NULL;
	
	p = phead;
	node * curTem;
	while(p != NULL){
		tem = p->next;
		p->next = cur;
		curTem = cur->next;
		//處理奇數個節點的情況
		if(tem == NULL && len % 2){
			cur->next = curTem;
		}else{
			cur->next = tem;
		}
		p = tem;
		cur = curTem;
		
	}
}

int main()
{
	int n;
	scanf("%d",&n);
	
	//輸入n個整數 
	input(n);
	
	//遍歷單鏈表
	show(head);
	
	//摺疊鏈表  1 2 3 4 5 變成 1 5 2 4 3 
	halfReverse(head);
	
	show(head);
	
	return 0;
}