[LeetCode] #206: Reverse Linked List (遞代&遞歸解法)

原題：

Reverse a singly linked list.

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

既然問了可否iteratively or recursively, 那就both把.

iterative 解法：

總結就是獲得下一個節點，更改當前節點指向，將指針往下移動，直到過完整個linkedlist.

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

recursive 解法：

總結是傳給helper method兩個節點，cur和pre（最開始是head和null), 先用n1存當前節點的next，而後把當前節點的next指向pre，而後一直recursively call help method直到過完整個linkedlist.

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ListNode reverseList(ListNode head) {
        if(head == null||head.next== null)
            return head;
        return getReverse(head, null);
    }
    
    public ListNode getReverse(ListNode cur, ListNode prev){
        if(cur.next == null){
            cur.next = prev;
            return cur;
        }
        ListNode n1 = cur.next;
        cur.next = prev;
        return getReverse(n1,cur);
    }
}