反转链表的各种问题

这道题，使用 迭代来做的话比较好理解一些，

迭代法：

class Solution {
    public ListNode reverseList(ListNode head) {
        if(head==null||head.next==null)
        	return head;
        ListNode newTemp = null;
        LitNode curNode = head;
        while(curNode!=null){
        	ListNode temp = curNode.next;
        	curNode.next = newTemp;
        	newTemp = curNode;
        	curNode = temp;
        }
        return newTemp;
    }
}

当然使用 递归法 会利于后续 反转链表的plus题

递归法：

class Solution{
    public ListNode reverseList(ListNode head){
        if(head==null||head.next==null)
            retrun head;
        ListNode last = reverseList(head.next);
        head.next.next = head;
        head.next = null;
        return last;
    }
}

递归法：

class Solution{
    ListNode beforeNode = null;
    public ListNode reverseList(ListNode head, int n){
        if(n==1){
            beforeNode = head.next;
            return head;
        }
        ListNode last = reverseList(head.next, n-1);
        head.next.next = head;
        head.next = beforeNode;
        return last;
    }
}

反转链表的 第 m ~ n个节点

class Solution{
    ListNode beforeNode = null;
    public ListNode reverseList(ListNode head, int m ,int n){
        if(m==1){
            return reverseList(head, n);
        }
        head.next = reverseList(head.next, m-1, n-1);
        return head;
    }

    private ListNode reverseList(ListNode head, int n){
        if(n==1){
            beforeNode = head.next;
            return head;
        }
        ListNode last = reverseList(head.next, n-1);
        head.next.next = head;
        head.next = beforeNode;
        return last;
    }
}