Single List Reversion

LeetCode

1. 基于头插法的迭代：

public ListNode reverseList(ListNode head) {
        if(head == null) return null;
        ListNode res = new ListNode(0);
        res.next = head;
        ListNode cur = head;
        while(cur.next != null){
            ListNode tmp = cur.next;
            cur.next = tmp.next;
            tmp.next = res.next;
            res.next = tmp;
        }
        return res.next;
}

2. 就地逆置的简洁迭代（elegant）

public ListNode reverse(ListNode head){
        ListNode prev = null;
        while(head != null){
            ListNode next = head.next;
            head.next = prev;
            prev = head;
            head = next;
        }
        return prev;
    }

3. 递归

ListNode* reverseList(ListNode* head) {
        if (!head || !(head -> next)) return head;
        ListNode* node = reverseList(head -> next);
        head -> next -> next = head;
        head -> next = NULL;
        return node; 
}