文章目录

• 思路
• 解题方法
• 复杂度
• Code

思路

双指针算法,定义pre,和cur,来遍历链表.然后进行反转

解题方法

定义一个temp记录cur->next的位置,当反转通过cur->next=pre来完成, 所以之后pre = cur, cur=temp(向后移动),直到cur为空,即达到遍历完成的效果

复杂度

• 时间复杂度:

  • O ( n )

• 空间复杂度:

  • O ( n )

Code

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public ListNode reverseList(ListNode head) {
        ListNode pre = null;
        ListNode cur = head;
        ListNode temp = new ListNode(0);

        while(cur != null){
            temp = cur.next; // 在cur.next改变之前
            cur.next = pre;
            pre = cur;
            cur = temp;
        }
        return pre;
    }
}

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
class Solution {
public:
    ListNode* reverseList(ListNode* head) {
        ListNode* cur = head;
        ListNode* pre = NULL;
        ListNode* temp = new ListNode(0);
        while(cur){
            temp = cur->next;
            cur->next = pre;
            pre = cur;
            cur = temp;
        }
        return pre;
    }
};