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Commit afd770d

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feat: update solutions to lc problems
1 parent d9e00c6 commit afd770d

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7 files changed

+54
-8
lines changed

7 files changed

+54
-8
lines changed

‎solution/0000-0099/0081.Search in Rotated Sorted Array II/README.md‎

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<!-- 这里可写通用的实现逻辑 -->
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二分查找。
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**方法一:二分查找**
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<!-- tabs:start -->
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‎solution/0000-0099/0083.Remove Duplicates from Sorted List/README.md‎

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# self.val = val
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# self.next = next
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class Solution:
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def deleteDuplicates(self, head: ListNode) -> ListNode:
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def deleteDuplicates(self, head: Optional[ListNode]) -> Optional[ListNode]:
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cur = head
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while cur and cur.next:
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if cur.val == cur.next.val:
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return head
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```
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```python
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# Definition for singly-linked list.
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# class ListNode:
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# def __init__(self, val=0, next=None):
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# self.val = val
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# self.next = next
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class Solution:
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def deleteDuplicates(self, head: Optional[ListNode]) -> Optional[ListNode]:
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dummy = ListNode(1000)
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cur = dummy
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while head:
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if head.val != cur.val:
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cur.next = head
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cur = cur.next
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head = head.next
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cur.next = None
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return dummy.next
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```
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### **Java**
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<!-- 这里可写当前语言的特殊实现逻辑 -->

‎solution/0000-0099/0083.Remove Duplicates from Sorted List/README_EN.md‎

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# self.val = val
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# self.next = next
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class Solution:
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def deleteDuplicates(self, head: ListNode) -> ListNode:
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def deleteDuplicates(self, head: Optional[ListNode]) -> Optional[ListNode]:
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cur = head
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while cur and cur.next:
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if cur.val == cur.next.val:
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return head
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```
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```python
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# Definition for singly-linked list.
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# class ListNode:
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# def __init__(self, val=0, next=None):
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# self.val = val
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# self.next = next
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class Solution:
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def deleteDuplicates(self, head: Optional[ListNode]) -> Optional[ListNode]:
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dummy = ListNode(1000)
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cur = dummy
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while head:
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if head.val != cur.val:
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cur.next = head
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cur = cur.next
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head = head.next
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cur.next = None
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return dummy.next
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```
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### **Java**
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```java

‎solution/0000-0099/0086.Partition List/README.md‎

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<!-- 这里可写通用的实现逻辑 -->
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**方法一:模拟**
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创建两个链表,一个存放小于 `x` 的节点,另一个存放大于等于 `x` 的节点,之后进行拼接即可。
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时间复杂度 $O(n),空间复杂度 $O(1)$。其中 $n$ 是原链表的长度。
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<!-- tabs:start -->
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### **Python3**
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# self.val = val
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# self.next = next
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class Solution:
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def partition(self, head: ListNode, x: int) -> ListNode:
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def partition(self, head: Optional[ListNode], x: int) -> Optional[ListNode]:
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d1, d2 = ListNode(), ListNode()
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t1, t2 = d1, d2
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while head:

‎solution/0000-0099/0086.Partition List/README_EN.md‎

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# self.val = val
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# self.next = next
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class Solution:
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def partition(self, head: ListNode, x: int) -> ListNode:
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def partition(self, head: Optional[ListNode], x: int) -> Optional[ListNode]:
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d1, d2 = ListNode(), ListNode()
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t1, t2 = d1, d2
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while head:

‎solution/0000-0099/0086.Partition List/Solution.py‎

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# self.val = val
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# self.next = next
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class Solution:
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def partition(self, head: ListNode, x: int) -> ListNode:
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def partition(self, head: Optional[ListNode], x: int) -> Optional[ListNode]:
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d1, d2 = ListNode(), ListNode()
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t1, t2 = d1, d2
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while head:

‎solution/0100-0199/0105.Construct Binary Tree from Preorder and Inorder Traversal/README.md‎

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<!-- 这里可写通用的实现逻辑 -->
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前序序列的第一个结点 `preorder[0]` 为根节点,我们在中序序列中找到根节点的位置 i,可以将中序序列划分为左子树 `inorder[:i]` 、右子树 `inorder[i+1:]`。
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**方法一:递归**
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通过左右子树的区间,可以计算出左、右子树节点的个数,假设为 m、n。然后在前序节点中,从根节点往后的 m 个节点为左子树,再往后的 n 个节点为右子树。
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前序序列的第一个结点 $preorder[0]$ 为根节点,我们在中序序列中找到根节点的位置 $i,ドル可以将中序序列划分为左子树 $inorder[0..i]$ 、右子树 $inorder[i+1..]$。
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通过左右子树的区间,可以计算出左、右子树节点的个数,假设为 $m$ 和 $n$。然后在前序节点中,从根节点往后的 $m$ 个节点为左子树,再往后的 $n$ 个节点为右子树。
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递归求解即可。
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> 前序遍历:先遍历根节点,再遍历左右子树;中序遍历:先遍历左子树,再遍历根节点,最后遍历右子树。
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时间复杂度 $O(n),ドル空间复杂度 $O(n)$。
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<!-- tabs:start -->
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### **Python3**

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