[剑指offer][JAVA]面试题第[24]题[反转链表][递归][双指针][头插法]

Dear 丶 2023-02-17 11:28 134阅读 0赞

#### 【问题描述】\[中等\] ####

定义一个函数，输入一个链表的头节点，反转该链表并输出反转后链表的头节点。
    
     
    
    示例:
    
    输入: 1->2->3->4->5->NULL
    输出: 5->4->3->2->1->NULL
     
    
    限制：
    
    0 <= 节点个数 <= 5000

#### 【解答思路】 ####

###### 1. 双指针/头插法 ######

![在这里插入图片描述][watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L2RhZG9uZ3d1ZGk_size_16_color_FFFFFF_t_70]  
![在这里插入图片描述][watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L2RhZG9uZ3d1ZGk_size_16_color_FFFFFF_t_70 1]  
![在这里插入图片描述][watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L2RhZG9uZ3d1ZGk_size_16_color_FFFFFF_t_70 2]  
![在这里插入图片描述][watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L2RhZG9uZ3d1ZGk_size_16_color_FFFFFF_t_70 3]  
**时间复杂度：O(N) 空间复杂度：O(1)**

class Solution { 
    	public ListNode reverseList(ListNode head) { 
    		//申请节点，pre和 cur，pre指向null
    		ListNode pre = null;
    		ListNode cur = head;
    		ListNode tmp = null;
    		while(cur!=null) { 
    			//记录当前节点的下一个节点
    			tmp = cur.next;
    			//然后将当前节点指向pre
    			cur.next = pre;
    			//pre和cur节点都前进一位
    			pre = cur;
    			cur = tmp;
    		}
    		return pre;
    	}
    }

###### 2. 递归 ######

![在这里插入图片描述][watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L2RhZG9uZ3d1ZGk_size_16_color_FFFFFF_t_70 4]  
![在这里插入图片描述][watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L2RhZG9uZ3d1ZGk_size_16_color_FFFFFF_t_70 5]  
![在这里插入图片描述][watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L2RhZG9uZ3d1ZGk_size_16_color_FFFFFF_t_70 6]  
![在这里插入图片描述][watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L2RhZG9uZ3d1ZGk_size_16_color_FFFFFF_t_70 7]  
**时间复杂度：O(N) 空间复杂度：O(1)**

class Solution { 
    	public ListNode reverseList(ListNode head) { 
    		//递归终止条件是当前为空，或者下一个节点为空
    		if(head==null || head.next==null) { 
    			return head;
    		}
    		//这里的cur就是最后一个节点
    		ListNode cur = reverseList(head.next);
    		//这里请配合动画演示理解
    		//如果链表是 1->2->3->4->5，那么此时的cur就是5
    		//而head是4，head的下一个是5，下下一个是空
    		//所以head.next.next 就是5->4
    		head.next.next = head;
    		//防止链表循环，需要将head.next设置为空
    		head.next = null;
    		//每层递归函数都返回cur，也就是最后一个节点
    		return cur;
    	}
    }

#### 【总结】 ####

###### 1.链表题目 递归和双指针思想非常重要 ######

###### 2. 链表题目一定要画图 ######

转载：https://leetcode-cn.com/problems/fan-zhuan-lian-biao-lcof/solution/ru-guo-ni-kan-wan-ping-lun-he-ti-jie-huan-you-wen-/

链接：https://leetcode-cn.com/problems/fan-zhuan-lian-biao-lcof/solution/dong-hua-yan-shi-duo-chong-jie-fa-206-fan-zhuan-li/

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