143. 重排链表

Author: chenpengcode

linkedlist/143_reorderList.py

@@ -0,0 +1,80 @@
+# Definition for singly-linked list.
+class ListNode:
+    def __init__(self, val=0, next=None):
+        self.val = val
+        self.next = next
+
+
+class Solution:
+    def reorderList(self, head: ListNode) -> None:
+        """
+        Do not return anything, modify head in-place instead.
+        """
+        if not head:
+            return
+        vec = list()
+
+        node = head
+        while node:
+            vec.append(node)
+            node = node.next
+
+        i, j = 0, len(vec) - 1
+
+        while i < j:
+            vec[i].next = vec[j]
+            i += 1
+            vec[j].next = vec[i]
+            j -= 1
+        vec[i].next = None
+
+    def reorderList_2(self, head: ListNode) -> None:
+        def find_mid(head: ListNode):
+            slow = fast = head
+            while fast.next and fast.next.next:
+                slow = slow.next
+                fast = fast.next.next
+            return slow
+
+        def reverse(head: ListNode):
+            pre = None
+            cur = head
+            while cur:
+                tmp = cur.next
+                cur.next = pre
+                pre = cur
+                cur = tmp
+            return pre
+
+        def merge(l1: ListNode, l2: ListNode):
+            while l1 and l2:
+                tmp1 = l1.next
+                tmp2 = l2.next
+
+                l1.next = l2
+                l1 = tmp1
+                l2.next = l1
+                l2 = tmp2
+
+        if not head:
+            return
+        mid = find_mid(head)
+        l1, l2 = head, mid.next
+        mid.next = None
+        l2 = reverse(l2)
+        merge(l1, l2)
+
+
+if __name__ == '__main__':
+    node1 = ListNode(1)
+    node2 = ListNode(2)
+    node3 = ListNode(3)
+    node4 = ListNode(4)
+    node5 = ListNode(5)
+    node1.next = node2
+    node2.next = node3
+    node3.next = node4
+    node4.next = node5
+
+    solution = Solution()
+    print(solution.reorderList_2(node1))