25 Reverse Nodes in k-Group

LeetCode

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

k is a positive integer and is less than or equal to the length of the linked 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 list's nodes, only nodes themselves may be changed.

Example 1:

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

Example 2:

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

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每次先进行判断，剩下的元素是否足够，再转换。时间复杂度 o(n) ，空间复杂度 o(k)

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public ListNode reverseKGroup(ListNode head, int k) {
        ListNode h = new ListNode(0,head);
        ListNode t = h;
        while(hasNext(t, k)){
            t = modify(t, k);
            for(int i=0;i<k;++i) t=t.next;
        }
        return h.next;
    }
    
    private ListNode modify(ListNode head, int k){
        Deque<ListNode> stack = new LinkedList<>();
        ListNode t = head.next;
        for(int i=0;i<k;++i){
            stack.push(t);
            t=t.next;
        }
        ListNode tail = t;
        t = head;
        while(!stack.isEmpty()){
            t.next = stack.pop();
            t=t.next;
        }
        t.next = tail;
        return head;
    }
    
    private boolean hasNext(ListNode head, int k){
        for(int i=0;i<k;++i){
            head=head.next;
            if(head==null) return false;
        }
        return true;
    }
}

空间复杂度 o(1) 的解法

public ListNode reverseKGroup(ListNode head, int k) {
    if(k<=1) return head;
    ListNode dummy=new ListNode();
    dummy.next=head;
    ListNode t=dummy,temp,a,b,c=null;
    while(true){
        temp=t;
        for(int i=0;i<k;++i){
            temp=temp.next;
            if(temp==null) return dummy.next;
        }
        a=temp.next;
        b=t.next;
        c=b;
        for(int i=0;i<k;++i){ 
            c=c.next;
            b.next=a;
            a=b;
            b=c;
            
        }
        temp=t.next;
        t.next=a;
        t=temp;
    }
}