add solutions for Intersection-of-Two-Linked-Lists

Intersection-of-Two-Linked-Lists/Solution.cpp

@@ -0,0 +1,23 @@
+#include <stdio.h>
+
+struct ListNode {
+    int val;
+    ListNode *next;
+    ListNode(int x) : val(x), next(NULL) {}
+};
+
+class Solution {
+public:
+    ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {
+        ListNode *a = headA, *b = headB, *lastA = a, *lastB = b;
+        for (; a != NULL && b != NULL; lastA = a, lastB = b, a = a->next, b = b->next);
+        for (; b != NULL; lastB = b, b = b->next, headB = headB->next);
+        for (; a != NULL; lastA = a, a = a->next, headA = headA->next);
+        for (; headA != NULL && headB != NULL && headA != headB; headA = headA->next, headB = headB->next);
+        return (lastA != lastB) ? NULL :headA;
+    }
+};
+
+int main(int argc, char *argv[]) {
+    return 0;
+}

Intersection-of-Two-Linked-Lists/Solution.java

@@ -0,0 +1,20 @@
+public class Solution {
+
+    public class ListNode {
+        int val;
+        ListNode next;
+        ListNode(int x) {
+            val = x;
+            next = null;
+        }
+    }
+
+    public ListNode getIntersectionNode(ListNode headA, ListNode headB) {
+        ListNode a = headA, b = headB, lastA = a, lastB = b;
+        for (; a != null && b != null; lastA = a, lastB = b, a = a.next, b = b.next);
+        for (; b != null; lastB = b, b = b.next, headB = headB.next);
+        for (; a != null; lastA = a, a = a.next, headA = headA.next);
+        for (; headA != null && headB != null && headA != headB; headA = headA.next, headB = headB.next);
+        return (lastA != lastB) ? null :headA;
+    }
+}

Intersection-of-Two-Linked-Lists/Solution.py

@@ -0,0 +1,19 @@
+# Definition for singly-linked list.
+# class ListNode:
+#     def __init__(self, x):
+#         self.val = x
+#         self.next = None
+
+class Solution:
+    # @param two ListNodes
+    # @return the intersected ListNode
+    def getIntersectionNode(self, headA, headB):
+        s = set()
+        while headA is not None:
+            s.add(headA)
+            headA = headA.next
+        while headB is not None:
+            if headB in s:
+                return headB
+            headB = headB.next
+        return None