feat: add python and java solutions to leetcode problem: No.0160

Author: yanglbme

README.md

@@ -65,6 +65,7 @@
 1. [重排链表](/solution/0100-0199/0143.Reorder%20List/README.md)
 1. [旋转链表](/solution/0000-0099/0061.Rotate%20List/README.md)
 1. [回文链表](/solution/0200-0299/0234.Palindrome%20Linked%20List/README.md)
+1. [相交链表](/solution/0100-0199/0160.Intersection%20of%20Two%20Linked%20Lists/README.md)
 1. [环形链表](/solution/0100-0199/0141.Linked%20List%20Cycle/README.md)
 1. [环形链表 II](/solution/0100-0199/0142.Linked%20List%20Cycle%20II/README.md)
 

README_EN.md

@@ -63,6 +63,7 @@ Complete solutions to [LeetCode](https://leetcode-cn.com/problemset/all/), [LCOF
 1. [Reorder List](/solution/0100-0199/0143.Reorder%20List/README_EN.md)
 1. [Rotate List](/solution/0000-0099/0061.Rotate%20List/README_EN.md)
 1. [Palindrome Linked List](/solution/0200-0299/0234.Palindrome%20Linked%20List/README_EN.md)
+1. [Intersection of Two Linked Lists](/solution/0100-0199/0160.Intersection%20of%20Two%20Linked%20Lists/README_EN.md)
 1. [Linked List Cycle](/solution/0100-0199/0141.Linked%20List%20Cycle/README_EN.md)
 1. [Linked List Cycle II](/solution/0100-0199/0142.Linked%20List%20Cycle%20II/README_EN.md)
 

solution/0100-0199/0160.Intersection of Two Linked Lists/README.md

@@ -62,22 +62,63 @@
 
 <!-- 这里可写通用的实现逻辑 -->
 
+定义两个指针，遍历时，当两个指针到达末尾的节点指向另一个链表的头部继续遍历，最后如果相遇则为交点(在第一轮移动中恰好抹除了长度差)。
+
+两个指针等于移动了相同的距离，有交点就返回，无交点就是各走了两条指针的长度。
+
 <!-- tabs:start -->
 
 ### **Python3**
 
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```python
-
+# Definition for singly-linked list.
+# class ListNode:
+#     def __init__(self, x):
+#         self.val = x
+#         self.next = None
+
+class Solution:
+    def getIntersectionNode(self, headA: ListNode, headB: ListNode) -> ListNode:
+        if headA is None or headB is None:
+            return None
+        pA, pB = headA, headB
+        while pA != pB:
+            pA = pA.next if pA else headB
+            pB = pB.next if pB else headA
+        return pA
 ```
 
 ### **Java**
 
 <!-- 这里可写当前语言的特殊实现逻辑 -->
 
 ```java
-
+/**
+ * Definition for singly-linked list.
+ * public class ListNode {
+ *     int val;
+ *     ListNode next;
+ *     ListNode(int x) {
+ *         val = x;
+ *         next = null;
+ *     }
+ * }
+ */
+public class Solution {
+    public ListNode getIntersectionNode(ListNode headA, ListNode headB) {
+        if (headA == null || headB == null) {
+            return null;
+        }
+        ListNode pA = headA, pB = headB;
+        while (pA != pB) {
+            pA = pA != null ? pA.next : headB;
+            pB = pB != null ? pB.next : headA;
+        }
+        return pA;
+    }
+}
 ```
 
 ### **...**

solution/0100-0199/0160.Intersection of Two Linked Lists/README_EN.md

@@ -78,13 +78,50 @@
 ### **Python3**
 
 ```python
-
+# Definition for singly-linked list.
+# class ListNode:
+#     def __init__(self, x):
+#         self.val = x
+#         self.next = None
+
+class Solution:
+    def getIntersectionNode(self, headA: ListNode, headB: ListNode) -> ListNode:
+        if headA is None or headB is None:
+            return None
+        pA, pB = headA, headB
+        while pA != pB:
+            pA = pA.next if pA else headB
+            pB = pB.next if pB else headA
+        return pA
 ```
 
 ### **Java**
 
 ```java
-
+/**
+ * Definition for singly-linked list.
+ * public class ListNode {
+ *     int val;
+ *     ListNode next;
+ *     ListNode(int x) {
+ *         val = x;
+ *         next = null;
+ *     }
+ * }
+ */
+public class Solution {
+    public ListNode getIntersectionNode(ListNode headA, ListNode headB) {
+        if (headA == null || headB == null) {
+            return null;
+        }
+        ListNode pA = headA, pB = headB;
+        while (pA != pB) {
+            pA = pA != null ? pA.next : headB;
+            pB = pB != null ? pB.next : headA;
+        }
+        return pA;
+    }
+}
 ```
 
 ### **...**

solution/0100-0199/0160.Intersection of Two Linked Lists/Solution.java

@@ -14,33 +14,11 @@ public ListNode getIntersectionNode(ListNode headA, ListNode headB) {
         if (headA == null || headB == null) {
             return null;
         }
-        int lenA = 0, lenB = 0;
-        ListNode p = headA, q = headB;
-        while (p != null) {
-            p = p.next;
-            ++lenA;
+        ListNode pA = headA, pB = headB;
+        while (pA != pB) {
+            pA = pA != null ? pA.next : headB;
+            pB = pB != null ? pB.next : headA;
         }
-        while (q != null) {
-            q = q.next;
-            ++lenB;
-        }
-        
-        p = headA;
-        q = headB;
-        if (lenA > lenB) {
-            for (int i = 0; i < lenA - lenB; ++i) {
-                p = p.next;
-            }
-        } else {
-            for (int i = 0; i < lenB - lenA; ++i) {
-                q = q.next;
-            }
-        }
-        while (p != null && p != q) {
-            p = p.next;
-            q = q.next;
-        }
-        return p;
-        
+        return pA;
     }
 }

solution/0100-0199/0160.Intersection of Two Linked Lists/Solution.py

@@ -0,0 +1,15 @@
+# Definition for singly-linked list.
+# class ListNode:
+#     def __init__(self, x):
+#         self.val = x
+#         self.next = None
+
+class Solution:
+    def getIntersectionNode(self, headA: ListNode, headB: ListNode) -> ListNode:
+        if headA is None or headB is None:
+            return None
+        pA, pB = headA, headB
+        while pA != pB:
+            pA = pA.next if pA else headB
+            pB = pB.next if pB else headA
+        return pA