add solution of problem 160: intersection of two linked lists

Author: ppulse

IntersectionOfTwoLinkedLists160/README.md

@@ -0,0 +1,19 @@
+Write a program to find the node at which the intersection of two singly linked lists begins.
+
+
+For example, the following two linked lists:
+
+A:          a1 → a2
+                   ↘
+                     c1 → c2 → c3
+                   ↗            
+B:     b1 → b2 → b3
+begin to intersect at node c1.
+
+
+Notes:
+
+- If the two linked lists have no intersection at all, return `null`.
+- The linked lists must retain their original structure after the function returns.
+- You may assume there are no cycles anywhere in the entire linked structure.
+- Your code should preferably run in O(n) time and use only O(1) memory.

IntersectionOfTwoLinkedLists160/Solution.java

@@ -0,0 +1,62 @@
+/**
+ * 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 currNodeA = headA;
+        ListNode currNodeB = headB;
+        
+        int lenA = 0;
+        while (currNodeA != null) {
+            currNodeA = currNodeA.next;
+            lenA++;
+        }
+        
+        int lenB = 0;
+        while (currNodeB != null) {
+            currNodeB = currNodeB.next;
+            lenB++;
+        }
+        
+        int diff = 0;
+        
+        if (lenA > lenB) {
+            currNodeA = headA;
+            currNodeB = headB;
+            diff = lenA - lenB;
+        } else {
+            currNodeA = headB;
+            currNodeB = headA;
+            diff = lenB - lenA;
+        }
+        
+        while (diff-- > 0) {
+            currNodeA = currNodeA.next;
+        }
+        
+        ListNode result = null;
+        
+        while (currNodeA != null && currNodeB != null) {
+            if (currNodeA != currNodeB) {
+                currNodeA = currNodeA.next;
+                currNodeB = currNodeB.next;
+            } else {
+                result = currNodeA;
+                break;
+            }
+        }
+        
+        return result;
+    }
+}