算法：翻转链表206. Reverse Linked List

Posted 2021-10-26 架构师易筋

206. Reverse Linked List

Given the head of a singly linked list, reverse the list, and return the reversed list.

Example 1:

Input: head = [1,2,3,4,5]
Output: [5,4,3,2,1]
Example 2:

Input: head = [1,2]
Output: [2,1]
Example 3:

Input: head = []
Output: []

Constraints:

The number of nodes in the list is the range [0, 5000].
-5000 <= Node.val <= 5000

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

遍历解法iterative solution

升维解决问题，增加两个变量，

1. 上一个节点ListNode newHead
2. 临时保存下一个节点ListNode next

public ListNode reverseList(ListNode head) {
    /* iterative solution */
    ListNode newHead = null;
    while (head != null) {
        ListNode next = head.next;
        head.next = newHead;
        newHead = head;
        head = next;
    }
    return newHead;
}

递归解法recursive solution

升维解决，增加一个子方法，参数变为2个。

public ListNode reverseList(ListNode head) {
    /* recursive solution */
    return reverseListInt(head, null);
}

private ListNode reverseListInt(ListNode head, ListNode newHead) {
    if (head == null)
        return newHead;
    ListNode next = head.next;
    head.next = newHead;
    return reverseListInt(next, head);
}