Reverse

206. Reverse Linked List

My Solutions:

Time: O(n); Space: O(1)

public ListNode reverseList(ListNode head) {
    if (head == null || head.next == null) return head;
    ListNode pre = null;
    while (head != null) {
        ListNode temp = head.next; // store next
        head.next = pre; // reverse current(head) node's pointer
        pre = head; // move pointers one position ahead
        head = temp;
    }
    return pre;
}

• recursive version

public ListNode reverseList(ListNode head) {
    if (head == null || head.next == null) return head;
    ListNode newHead = reverseList(head.next);
    head.next.next = head;
    head.next = null;
    return newHead;
}

92. Reverse Linked List II

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.

My Solutions:

只反转从m到n位置的nodes，所以需要先走到m之前的位置，然后反转到n-m的位置

Time: O(n); Space: O(1)

class Solution {
    public ListNode reverseBetween(ListNode head, int m, int n) {
        if (head == null || m > n) return null;

        ListNode dummy = new ListNode(0);
        dummy.next = head;
        ListNode pre = dummy;
        
        //find the node right before the reversed one
        for (int i = 0; i < m - 1; i++) pre = pre.next; 

        ListNode first = pre.next;
        ListNode second = first.next;

        for (int i = 0; i < n - m; i++) {
            first.next = second.next;
            second.next = pre.next;
            pre.next = second;
            second = first.next;
        }

        return dummy.next;
    }
}

25。Reverse Nodes in k-Group

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]

My Solutions:

这道题和#92是类似的。

public ListNode reverseKGroup(ListNode head, int k) {
    if (head == null || k == 1) return head;
    ListNode dummy = new ListNode(0);
    dummy.next = head;

    ListNode curr = dummy;
    int count = 0; // count of nodes in the list
    while (curr.next != null) {
        curr = curr.next;
        count++;
    }

    // dummy -->  1  -->  2  --> 3 --> 4
    //  pre      curr   next
    
    // dummy --> 2 --> 1  --> 3 --> 4
    // pre           curr   next

    // dummy --> 3 --> 2  --> 1 --> 4
    // pre      next         curr
    // pre                   curr  next
    
    // dummy --> 3 --> 2  --> 1 --> 4 --> 5
    //                       pre
    //                            curr
    //                                   next

    ListNode pre = dummy, next = dummy;
    // do the reverse unless there're not enough nodes
    while (count >= k) {
        curr = pre.next; // curr is 1
        next = curr.next; // next is 2
        for (int i = 1; i < k; i++) {
            curr.next = next.next;  // connect 1 to 3           // connect 1 to 4
            next.next = pre.next;   // connect 2 to 1           // connect 3 to 2
            pre.next = next;        // connect dummy to 2       // connect dummy to 3
            next = curr.next;       // put next pointer to 1    // put next pointer to 1
        }
        pre = curr;
        count -= k;
    }
    return dummy.next;
}