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Author: NinePiece2

Solutions/0328-odd-even-linked-list/README.md

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+<p>Given the <code>head</code> of a singly linked list, group all the nodes with odd indices together followed by the nodes with even indices, and return <em>the reordered list</em>.</p>
+
+<p>The <strong>first</strong> node is considered <strong>odd</strong>, and the <strong>second</strong> node is <strong>even</strong>, and so on.</p>
+
+<p>Note that the relative order inside both the even and odd groups should remain as it was in the input.</p>
+
+<p>You must solve the problem&nbsp;in <code>O(1)</code>&nbsp;extra space complexity and <code>O(n)</code> time complexity.</p>
+
+<p>&nbsp;</p>
+<p><strong class="example">Example 1:</strong></p>
+<img alt="" src="https://assets.leetcode.com/uploads/2021/03/10/oddeven-linked-list.jpg" style="width: 300px; height: 123px;" />
+<pre>
+<strong>Input:</strong> head = [1,2,3,4,5]
+<strong>Output:</strong> [1,3,5,2,4]
+</pre>
+
+<p><strong class="example">Example 2:</strong></p>
+<img alt="" src="https://assets.leetcode.com/uploads/2021/03/10/oddeven2-linked-list.jpg" style="width: 500px; height: 142px;" />
+<pre>
+<strong>Input:</strong> head = [2,1,3,5,6,4,7]
+<strong>Output:</strong> [2,3,6,7,1,5,4]
+</pre>
+
+<p>&nbsp;</p>
+<p><strong>Constraints:</strong></p>
+
+<ul>
+	<li>The number of nodes in the linked list is in the range <code>[0, 10<sup>4</sup>]</code>.</li>
+	<li><code>-10<sup>6</sup> &lt;= Node.val &lt;= 10<sup>6</sup></code></li>
+</ul>

Solutions/0328-odd-even-linked-list/solution.py

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+# Definition for singly-linked list.
+# class ListNode:
+#     def __init__(self, val=0, next=None):
+#         self.val = val
+#         self.next = next
+class Solution:
+    def oddEvenList(self, head: Optional[ListNode]) -> Optional[ListNode]:
+        if head is None:
+            return head
+        odd = head
+        even = total = head.next
+
+        while even and even.next:
+            odd.next = even.next
+            odd = odd.next
+            even.next = odd.next
+            even = even.next
+        
+        odd.next = total
+        return head