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

RobinQu 發(fā)布于2019-08-14 15:21 / 2904人閱讀

摘要：原題既然問了能否那就把解法總結(jié)就是得到下一個節(jié)點，更改當前節(jié)點指向，將指針往下移動，直到過完整個解法總結(jié)是傳給兩個節(jié)點，和最開始是和先用存當前節(jié)點的，然后把當前節(jié)點的指向，然后一直直到過完整個

原題：

Reverse a singly linked list.

click to show more hints.

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

既然問了能否iteratively or recursively, 那就both把.

iterative 解法：

總結(jié)就是得到下一個節(jié)點，更改當前節(jié)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(jié)是傳給helper method兩個節(jié)點，cur和pre（最開始是head和null), 先用n1存當前節(jié)點的next，然后把當前節(jié)點的next指向pre，然后一直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);
    }
}