Leetcode top 100 raised questions 25. Reverse nodes in k-group (Java version; Hard)

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Leetcode top 100 raised questions 25. Reverse nodes in k-group (Java version; Hard)

Title Description

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

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

Note:

Only constant extra memory is allowed.
You may not alter the values in the list's nodes, only nodes itself may be changed.

The first time; recursion, the core is recursive function logic: start from the head, reverse k nodes, and connect with the next section (the new condition hidden in this sentence is new recursion), and finally return to the head node

//This problem is very recursive; each section is processed in the same way, but how to write it? What is the logic of recursive function?
class Solution {
    //Recursive function logic: start from head, reverse k nodes, and connect with the next section (New recursion hidden in this sentence), and finally return to the head node
    public ListNode reverseKGroup(ListNode head, int k) {
        //base case
        if(head==null || head.next==null || k<=1)
            return head;
        //
        //Determine whether there are k nodes
        int n = 0;
        ListNode cur=head;
        while(cur!=null){
            n++;
            if(n==k)
                break;
            cur = cur.next;
        }
        //If it is less than k nodes, it does not need to flip and directly returns to the head node
        if(n<k)
            return head;
        //Length enough k, reverse the list
        //First save the head node of the next link list to be reversed
        ListNode nextHead = cur.next;
        ListNode left=null, right;
        cur = head;
        for(int i=0; i<k; i++){
            //save next
            right = cur.next;
            //change
            cur.next = left;
            //update
            left = cur;
            cur = right;
        }
        //The tail of the reversed linked list connects the new head of the next link to be reversed
        head.next = reverseKGroup(nextHead, k);
        //Returns the header of the current segment
        return left;
    }
}

Do it for the first time; create a dummy node so you don't have to deal with it alone; be concise

class Solution {
    public ListNode reverseKGroup(ListNode head, int k) {
        //input check
        if(head==null || head.next == null || k<=1)
            return head;
        //Create dummy nodes; treat dummy nodes as processed nodes
        ListNode dummy = new ListNode(0);
        dummy.next = head;
        //You need to know the tail after the previous reversal and the head of the next link list to be reversed
        ListNode preHead = dummy, preTail = dummy, nextHead = head;
        while(nextHead!=null){
            //Prepare to reverse the link list at the beginning of nextHead
            ListNode cur = nextHead;
            //Check whether the length is not less than k
            int n = 0;
            while(cur!=null){
                cur = cur.next;
                n++;
                if(n==k)
                    break;
            }
            //If the length is less than k, there is no need to reverse
            if(n<k)
                break;
            //Length not less than k, reverse operation
            ListNode left=null, right;
            cur = nextHead;
            for(int i=0; i<k; i++){
                //save next
                right = cur.next;
                //change
                cur.next = left;
                //update
                left = cur;
                cur = right;
            }
            //update
            //The tail of the previous section connects the head of the current section
            preTail.next = left;
            preTail = nextHead;
            nextHead = cur;
            //Details: for the time being, if you jump out of the loop, you don't need to connect again
            preTail.next = nextHead;
        }
        return dummy.next;
        
    }
}

Do it for the first time; draw a picture; deal with it alone first, and then deal with the rest in a circular way; write it a little verbose, especially if it needs to be dealt with separately

class Solution {
    public ListNode reverseKGroup(ListNode head, int k) {
        //input check
        if(head==null || head.next==null)
            return head;
        //Length of statistical link list
        int n=0;
        ListNode cur=head;
        while(cur!=null){
            n++;
            cur = cur.next;
        }
        if(n<k)
            return head;
        //Number of reversals
        int times = n/k;
        //Single reverse first
        ListNode[] res = reverse(head, k);
        //Head to return
        ListNode newHead = res[0];
        ListNode tail = res[1];
        ListNode nextHead = res[2];
        //Deal with the rest
        for(int i=1; i<times; i++){
            ListNode[] tmp = reverse(nextHead, k);
            //Old tail connects new head
            tail.next = tmp[0];
            //update
            tail = tmp[1];
            nextHead = tmp[2];
        }
        tail.next = nextHead;
        return newHead;
        
        
    }
    
    //Return the reversed: 1. Head node 2. Tail node 3. Next node of the original tail node
    private ListNode[] reverse(ListNode head, int k){
        ListNode nextHead = head;
        for(int i=0; i<k; i++)
            nextHead = nextHead.next;
        //
        ListNode left=null, cur=head, right;
        while(k>0){
            //save next
            right = cur.next;
            //change
            cur.next = left;
            //update
            left = cur;
            cur = right;
            k--;
        }
        //1. 2. 3.
        return new ListNode[]{left, head, nextHead};
    }
}

Excellent problem solving Recursive writing

class Solution {
    public ListNode reverseKGroup(ListNode head, int k) {
        ListNode cur = head;
        int count = 0;
        while (cur != null && count != k) {
            cur = cur.next;
            count++;
        }
        if (count == k) {
            cur = reverseKGroup(cur, k);
            while (count != 0) {
                count--;
                ListNode tmp = head.next;
                head.next = cur;
                cur = head;
                head = tmp;
            }
            head = cur;
        }
        return head;
    }
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Posted on Sat, 08 Feb 2020 07:16:57 -0800 by dimitris