206. Reverse Linked List

# Easy

Key idea: record pre-node + current node + change next direction

核心： 记录前一个节点 ＋ 记录当前节点 ＋ 改变指针方向

process of reverse a linked list

Solution

temp records next node, head is current node, preNode is the pre-node of head. Don't forget to initialise preNode = NULL, return new head.

temp 记录下一个节点，head表示当前节点，preNode表示head的母节点。必须初始化preNode=NULL，并且最后返回新的头节点（即最后一个节点）。

1. record next node temporally

2. let head.next point to preNode

3. move to next node

4. 暂时记录下一个节点

5. 让head.next指向前一个节点

6. 指针后移一位

Python syntax

Java

class Solution {
    public ListNode reverseList(ListNode head) {
        if(head == null || head.next == null) return head;
        ListNode pre = null;
        ListNode curr = new ListNode();
        curr = head;
        while(curr != null) {
            ListNode nextTemp = curr.next;
            curr.next = pre;
            pre = curr;
            curr = nextTemp;
        }
        return pre;
    }
}

Second Tab

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def reverseList(self, head: ListNode) -> ListNode:
        pre = None
        
        while head != None:
            temp = head.next
            head.next = pre
            
            pre = head
            head = temp
            
        return pre