143. Reorder List

143. Reorder List

Medium

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Given a singly linked list L: L0→L1→…→Ln-1→Ln,
reorder it to: L0→Ln→L1→Ln-1→L2→Ln-2→…

You may not modify the values in the list's nodes, only nodes itself may be changed.

Example 1:

Given 1->2->3->4, reorder it to 1->4->2->3.

Example 2:

Given 1->2->3->4->5, reorder it to 1->5->2->4->3.

Solution 1: leverage slow-fast two pointer to find the middle place, then split the list into two, reverse the second part, then insert it into the first part.

Solution 2: use a stack to store the list

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public void reorderList(ListNode head) {
        if (null == head || null == head.next) {
            return;
        }
        
        ListNode ptr = head;
        Stack stack = new Stack<>();
        while (ptr != null) {
            stack.push(ptr);
            ptr = ptr.next;
        }
        
        ptr = head;
        int halfOfStack = (stack.size() + 1) / 2;
        for (int i = 0; i < halfOfStack; i++) {
            ListNode tNode = stack.pop();
            tNode.next = ptr.next;
            ptr.next = tNode;
            
            ptr = tNode.next;
        }
        ptr.next = null;
        stack.clear();
    }
}