Description

Reverse a singly linked list.

Hint:A linked list can be reversed either iteratively or recursively. Could you implement both?

Solution

Iterative

可以用head作為curr，這樣可以省略一個(gè)局部變量。
另外prev初始為null是比較簡(jiǎn)潔的寫法。

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode reverseList(ListNode head) {
        ListNode prev = null;
        while (head != null) {
            ListNode next = head.next;
            head.next = prev;
            prev = head;
            head = next;
        }
        return prev;
    }
}

Recursive

遞歸的寫法還蠻tricky的，重點(diǎn)是保證reverse后原鏈表的head指向reverse后的鏈表尾部，而不是指向null。舉個(gè)例子：

• 原鏈表：1->2->3
• 一層遞歸后：1->2<-3
• 二層遞歸后：1<-2<-3

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

        ListNode newHead = reverseList(head.next);
        head.next.next = head;
        head.next = null;
        return newHead;
    }
}