[LeetCode]--143. Reorder List(Python + Java)

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

You must do this in-place without altering the nodes’ values.

For example,
Given {1,2,3,4}, reorder it to {1,4,2,3}.
Let’s give the code directly.

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public void reorderList(ListNode head) {
        if (head == null || head.next == null) return;

        // Find the middle of the list
        ListNode p1 = head;
        ListNode p2 = head;
        while (p2.next != null && p2.next.next != null) {
            p1 = p1.next;
            p2 = p2.next.next;
        }

        // Reverse the half after middle
        ListNode preMiddle = p1;
        ListNode preCurrent = p1.next;
        while (preCurrent.next != null) {
            ListNode current = preCurrent.next;
            preCurrent.next = current.next;
            current.next = preMiddle.next;
            preMiddle.next = current;
        }

        // Start reorder one by one.
        p1 = head;
        p2 = preMiddle.next;
        while (p1 != preMiddle) {
            preMiddle.next = p2.next;
            p2.next = p1.next;
            p1.next = p2;
            p1 = p2.next;
            p2 = preMiddle.next;
        }
    }
}

Let’s see some Python Code.

# Definition for singly-linked list.
# class ListNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution(object):
    def reorderList(self, head):
        if not head:
            return

        # find the mid point
        slow = fast = head 
        while fast and fast.next:
            slow = slow.next
            fast = fast.next.next

        # reverse the second half in-place
        pre, node = None, slow
        while node:
            pre, node.next, node = node, pre, node.next

        # Merge in-place; Note : the last node of "first" and "second" are the same
        first, second = head, pre
        while second.next:
            first.next, first = second, first.next
            second.next, second = first, second.next
        return 

Just the same idea!

Reference!
[1] All solution
[2] Python solution
[3] Java solution

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