真的是太久太久沒有刷題了。。。。那天阿裏面到這麽簡單的題目發現自己都寫不利索了。。。哭瞎。。。

Reverse a singly linked list.

可以用遞歸和非遞歸的方法。

遞歸：

主要思想是定義一個nextNode，用nextNode作為尾部直接連前面一個。

public class Solution {
    public ListNode reverseList(ListNode head) {
        if (head == null || head.next == null) {
            return head;
        }
        ListNode nextNode = head.next;
        ListNode newHead = reverseList(nextNode);
        nextNode.next = head;
        head.next = null;
        return newHead;
    }
}

非遞歸：

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

C++做法，雖然現在還不懂這個ListNode*是什麽意思。。。。

遞歸：

/**
 * 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* node = head->next;
        ListNode* newHead = reverseList(node);
        node -> next = head;
        head -> next = NULL;
        return newHead;
    }
};

非遞歸：

class Solution {
public:
    ListNode* reverseList(ListNode* head) {
        if (head == NULL || head -> next == NULL) {
            return head;
        }
        ListNode* pre = NULL;
        while (head != NULL) {
            ListNode* next = head -> next;
            head -> next = pre;
            pre = head;
            head = next;
        }
        return pre;
    }
};

Reverse Linked List Leetcode