LeetCode -- Reverse Nodes in k-Group

LeetCode -- Reverse Nodes in k-Group

題目描述：


Given a linked list, reverse the nodes of a linked list k at a time and return its modified list.


If the number of nodes is not a multiple of k then left-out nodes in the end should remain as it is.


You may not alter the values in the nodes, only nodes itself may be changed.


Only constant memory is allowed.


For example,
Given this linked list: 1->2->3->4->5


For k = 2, you should return: 2->1->4->3->5


For k = 3, you should return: 3->2->1->4->5


就是在遍歷鏈表的過程中，如果從當前節點開始算，節點數大於k個，反轉這k個節點；然後從當前位置向後移動k個位置，繼續反轉的過程，直到從當前節點到最後的節點數小於K個停止反轉。


思路：
1. 創建反轉函數，反轉從當前開始的k個節點。 這個函數空間復雜度O（n）
2. 先得到鏈表總長度，如果小於k直接返回，否則每次旋轉k個節點並指向下一次旋轉的位置。




實現代碼：

/**
 * 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 ReverseKGroup(ListNode head, int k) 
    {
        if(head == null || head.next == null || k == 1){
    		return head;
    	}
    	
    	var len = 0;
    	var h = head;
    	while(head != null){
    		head = head.next;
    		len ++;
    	}
    	
    	if(k > len){
    		return h;
    	}
    	
    	ListNode pre = null;
    	ListNode newHead = null;
    	while(len >= k)
    	{
    		ReverseKNodes(ref h, k, ref pre);
    		
    		if(pre == null){
    			newHead = h;
    			pre = h;
    			for(var i =0 ;i < k-1; i++){
    				pre = pre.next;	
    			}
    		}else{
    			for(var i = 0;i < k;i++){
    				pre = pre.next;
    			}
    		}
    		
    		for(var i = 0;i < k;i++){
    			h = h.next;
    		}
    		
    		len -= k;
    	}
    	
    	return newHead;
    }


private void ReverseKNodes(ref ListNode n, int k, ref ListNode preNode)
{
	var stack = new Stack();
	for(var i = 0;i < k; i++){
		stack.Push(n.val);
		n = n.next;
	}
	ListNode end = n;
	
	ListNode tmp = new ListNode(stack.Pop());
	var tmpHead = tmp;
	while(stack.Count > 0){
		var n1 = new ListNode(stack.Pop());
		tmp.next = n1;
		tmp = tmp.next;
	}
	
	tmp.next = end;
	
	if(preNode != null){
		n = tmpHead;
		preNode.next = tmpHead;
	}else{
		n = tmpHead;
	}
}
}