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feat: add python and java solutions to lcci question
添加《程序员面试金典》题解：面试题 02.07. 链表相交
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Diff for: ‎index.html

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<script>
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window.$docsify = {
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name: 'leetcode',
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// repo: 'doocs/leetcode',
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search: [
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'/','/solution/','/lcof/','lcci/'
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],

Diff for: ‎lcci/02.07.Intersection of Two Linked Lists/README.md

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## 解法
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<!-- 这里可写通用的实现逻辑 -->
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先求两链表的长度差 `differ`，接着较长的链表先走 `differ` 步，之后两链表同时走，若相遇，则说明相交。
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### Python3
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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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, x):
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# self.val = x
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# self.next = None
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class Solution:
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def getIntersectionNode(self, headA: ListNode, headB: ListNode) -> ListNode:
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len1, len2 = self._length(headA), self._length(headB)
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if len1 < len2:
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headA, headB = headB, headA
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differ = abs(len1 - len2)
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for _ in range(differ):
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headA = headA.next
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while headA:
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if headA == headB:
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return headA
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headA = headA.next
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headB = headB.next
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return None
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def _length(self, node: ListNode) -> int:
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n = 0
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while node:
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node = node.next
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n += 1
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return n
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```
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### Java

Diff for: ‎lcci/02.07.Intersection of Two Linked Lists/README_EN.md

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# [02.07. Intersection of Two Linked Lists](https://leetcode-cn.com/problems/intersection-of-two-linked-lists-lcci)
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## Description
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<p>Given two (singly) linked lists, determine if the two lists intersect. Return the inter&shy; secting node. Note that the intersection is defined based on reference, not value. That is, if the kth node of the first linked list is the exact same node (by reference) as the jth node of the second linked list, then they are intersecting.</p>
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<p><strong>Example 1: </strong></p>
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<pre>
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<strong>Input: </strong>intersectVal = 8, listA = [4,1,8,4,5], listB = [5,0,1,8,4,5], skipA = 2, skipB = 3
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<strong>Output: </strong>Reference of the node with value = 8
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<strong>Input Explanation:</strong> The intersected node&#39;s value is 8 (note that this must not be 0 if the two lists intersect). From the head of A, it reads as [4,1,8,4,5]. From the head of B, it reads as [5,0,1,8,4,5]. There are 2 nodes before the intersected node in A; There are 3 nodes before the intersected node in B.</pre>
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<p><strong>Example 2: </strong></p>
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<pre>
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<strong>Input: </strong>intersectVal = 2, listA = [0,9,1,2,4], listB = [3,2,4], skipA = 3, skipB = 1
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<strong>Output: </strong>Reference of the node with value = 2
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<strong>Input Explanation:</strong>&nbsp;The intersected node&#39;s value is 2 (note that this must not be 0 if the two lists intersect). From the head of A, it reads as [0,9,1,2,4]. From the head of B, it reads as [3,2,4]. There are 3 nodes before the intersected node in A; There are 1 node before the intersected node in B.</pre>
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<p><strong>Example 3: </strong></p>
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<pre>
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<strong>Input: </strong>intersectVal = 0, listA = [2,6,4], listB = [1,5], skipA = 3, skipB = 2
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<strong>Output: </strong>null
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<strong>Input Explanation:</strong> From the head of A, it reads as [2,6,4]. From the head of B, it reads as [1,5]. Since the two lists do not intersect, intersectVal must be 0, while skipA and skipB can be arbitrary values.
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<strong>Explanation:</strong> The two lists do not intersect, so return null.</pre>
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<p><b>Notes:</b></p>
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<ul>
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<li>If the two linked lists have no intersection at all, return&nbsp;<code>null</code>.</li>
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<li>The linked lists must retain their original structure after the function returns.</li>
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<li>You may assume there are no cycles anywhere in the entire linked structure.</li>
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<li>Your code should preferably run in O(n) time and use only O(1) memory.</li>
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</ul>
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## Solutions
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### Python3
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```python
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```
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### Java
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```java
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```
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### ...
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```
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```
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# [02.07. Intersection of Two Linked Lists](https://leetcode-cn.com/problems/intersection-of-two-linked-lists-lcci)
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## Description
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<p>Given two (singly) linked lists, determine if the two lists intersect. Return the inter&shy; secting node. Note that the intersection is defined based on reference, not value. That is, if the kth node of the first linked list is the exact same node (by reference) as the jth node of the second linked list, then they are intersecting.</p>
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<p><strong>Example 1: </strong></p>
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<pre>
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<strong>Input: </strong>intersectVal = 8, listA = [4,1,8,4,5], listB = [5,0,1,8,4,5], skipA = 2, skipB = 3
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<strong>Output: </strong>Reference of the node with value = 8
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<strong>Input Explanation:</strong> The intersected node&#39;s value is 8 (note that this must not be 0 if the two lists intersect). From the head of A, it reads as [4,1,8,4,5]. From the head of B, it reads as [5,0,1,8,4,5]. There are 2 nodes before the intersected node in A; There are 3 nodes before the intersected node in B.</pre>
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<p><strong>Example 2: </strong></p>
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<pre>
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<strong>Input: </strong>intersectVal = 2, listA = [0,9,1,2,4], listB = [3,2,4], skipA = 3, skipB = 1
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<strong>Output: </strong>Reference of the node with value = 2
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<strong>Input Explanation:</strong>&nbsp;The intersected node&#39;s value is 2 (note that this must not be 0 if the two lists intersect). From the head of A, it reads as [0,9,1,2,4]. From the head of B, it reads as [3,2,4]. There are 3 nodes before the intersected node in A; There are 1 node before the intersected node in B.</pre>
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<p><strong>Example 3: </strong></p>
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<pre>
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<strong>Input: </strong>intersectVal = 0, listA = [2,6,4], listB = [1,5], skipA = 3, skipB = 2
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<strong>Output: </strong>null
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<strong>Input Explanation:</strong> From the head of A, it reads as [2,6,4]. From the head of B, it reads as [1,5]. Since the two lists do not intersect, intersectVal must be 0, while skipA and skipB can be arbitrary values.
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<strong>Explanation:</strong> The two lists do not intersect, so return null.</pre>
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<p><b>Notes:</b></p>
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<ul>
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<li>If the two linked lists have no intersection at all, return&nbsp;<code>null</code>.</li>
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<li>The linked lists must retain their original structure after the function returns.</li>
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<li>You may assume there are no cycles anywhere in the entire linked structure.</li>
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<li>Your code should preferably run in O(n) time and use only O(1) memory.</li>
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</ul>
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## Solutions
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### Python3
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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, x):
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# self.val = x
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# self.next = None
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class Solution:
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def getIntersectionNode(self, headA: ListNode, headB: ListNode) -> ListNode:
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len1, len2 = self._length(headA), self._length(headB)
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if len1 < len2:
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headA, headB = headB, headA
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differ = abs(len1 - len2)
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for _ in range(differ):
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headA = headA.next
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while headA:
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if headA == headB:
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return headA
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headA = headA.next
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headB = headB.next
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return None
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def _length(self, node: ListNode) -> int:
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n = 0
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while node:
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node = node.next
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n += 1
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return n
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```
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### Java
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```java
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```
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### ...
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```
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```

Diff for: ‎lcci/02.07.Intersection of Two Linked Lists/Solution.py

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# Definition for singly-linked list.
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# class ListNode:
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# def __init__(self, x):
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# self.val = x
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# self.next = None
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class Solution:
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def getIntersectionNode(self, headA: ListNode, headB: ListNode) -> ListNode:
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len1, len2 = self._length(headA), self._length(headB)
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if len1 < len2:
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headA, headB = headB, headA
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differ = abs(len1 - len2)
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for _ in range(differ):
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headA = headA.next
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while headA:
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if headA == headB:
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return headA
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headA = headA.next
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headB = headB.next
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return None
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def _length(self, node: ListNode) -> int:
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n = 0
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while node:
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node = node.next
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n += 1
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return n

Diff for: ‎lcof/README_EN.md

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Readers will improve their interview performance after reading this book. It will be beneficial for them even after they get offers, because its topics, such as approaches to analyzing difficult problems, writing robust code and optimizing, are all essential for high-performing coders.
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[中文文档](/solution/README.md)
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[中文文档](/lcof/README.md)
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## Solutions
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English Solutions is not available for the book. Please switch to Chinese version.

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