当前位置：首页 > news >正文

LeetCode-- Reverse Linked List II

news 来源：原创 2024/5/20 12:05:36

题目描述：

Reverse a linked list from position m to n. Do it in-place and in one-pass.

For example:
Given 1->2->3->4->5->NULL, m = 2 and n = 4,

return 1->4->3->2->5->NULL.

Note:
Given m, n satisfy the following condition:
1 ≤ m ≤ n ≤ length of list.

思路：
1. 使用栈来保存m到n之间的数字，其余元素使用队列保存
2. 在[m,n]区间外时，循环弹出栈内元素到链表
3. 在[m,n]区间内，先循环弹出队列元素到链表，再创建栈，最后需要判断当前head是否为空

实现代码：


/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     public int val;
 *     public ListNode next;
 *     public ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ListNode ReverseBetween(ListNode head, int m, int n) {
        var stack = new Stack<int>();
        var q = new Queue<int>();
        ListNode node = null;
        var c = 1;
		ListNode newNode = null;
        while(head != null)
        {
            if(c >= m && c <=n){
                while(q.Count > 0){
                    var first = MoveNext(ref node , q.Dequeue());
					if(first){
						newNode = node;
					}
                }
                while(c >= m && c<= n){
                    stack.Push(head.val);
                    head = head.next;
                    c++;
                }
				if(head == null){
					 while(stack.Count > 0){
						var first = MoveNext(ref node , stack.Pop());
						if(first){
							newNode = node;
						}
					}
				}
            }
            else{
                while(stack.Count > 0){
                    var first = MoveNext(ref node , stack.Pop());
					if(first){
						newNode = node;
					}
                }
                var f = MoveNext(ref node , head.val);
				if(f)
				{
					newNode = node;
				}
                head = head.next;
				c++;
            }
        }
        
        return newNode;
    }
    
    private bool MoveNext(ref ListNode n , int val){
        if(n == null){
            n = new ListNode(val);
			return true;
        }
        else{
            n.next = new ListNode(val);
            n = n.next;
			return false;
        }
    }
    
}