【18】206. Reverse Linked List

206. Reverse Linked List

• Total Accepted: 188250
• Total Submissions: 429719
• Difficulty: Easy
• Contributors: Admin

Reverse a singly linked list.

Solution 1: Iterative 

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* reverseList(ListNode* head) {
        if(head == NULL || head -> next == NULL) return head;
        ListNode* dummy = new ListNode(-1);
        dummy -> next = head;
        /*time limitation
        ListNode* next = head -> next;
        while(next){//head节点是不变的，所以用next节点来判断
            dummy -> next = next;//把next节点一直连在开头
            head -> next = next -> next;
            next -> next = dummy -> next;
            next = head -> next;//next节点一直待在head后面
        }*/
        
        ListNode* cur = head;
        while(cur -> next){
            ListNode* tmp = cur -> next;
            cur -> next = tmp -> next;
            tmp -> next = dummy -> next;
            dummy -> next = tmp;
        }
        
        return dummy -> next;
    }
};

Solution 2: Recursive

递归解法的思路是，首先进入递归函数 不断递归，直到head指向最后一个节点。而p指向之前一个节点，调换head和p的位置，然后返回上一层递归函数，再交换p和head的位置，每次交换后，head节点后面都是交换好的顺序，直到p为首节点，然后再交换，首节点就成了为节点，此时整个链表也完成了翻转。

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* reverseList(ListNode* head) {
        if(head == NULL || head -> next == NULL) return head;
        ListNode* p = head;
        head = reverseList(p -> next);
        p -> next -> next = p;
        p -> next = NULL;
        return head;
    }
};