Linked List Cycle I && II

Change-Id: I66ec88cd863da9ebda61aae1e080a942cf90aff2

Author: applewjg

LinkedListCycle.java

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+/*
+ Author:     King, [email protected]
+ Date:       Oct 5, 2014
+ Problem:    Linked List Cycle
+ Difficulty: Easy
+ Source:     http://oj.leetcode.com/problems/linked-list-cycle/
+ Notes:
+ Given a linked list, determine if it has a cycle in it.
+ Follow up:
+ Can you solve it without using extra space?
+
+ Solution: two pointers.
+*/
+
+/**
+ * Definition for singly-linked list.
+ * class ListNode {
+ *     int val;
+ *     ListNode next;
+ *     ListNode(int x) {
+ *         val = x;
+ *         next = null;
+ *     }
+ * }
+ */
+public class Solution {
+    public boolean hasCycle(ListNode head) {
+        if (head == null || head.next == null) return false;
+        ListNode slow = head, fast = head;
+        while (fast != null && fast.next != null) {
+            slow = slow.next;
+            fast = fast.next.next;
+            if (slow == fast) break;
+        }
+        if (fast == null || fast.next == null) return false;
+        return true;
+    }
+}

LinkedListCycleII.java

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+/*
+ Author:     King, [email protected]
+ Date:       Jan 02, 2015
+ Problem:    Linked List Cycle II
+ Difficulty: Easy
+ Source:     http://oj.leetcode.com/problems/linked-list-cycle-ii/
+ Notes:
+ Given a linked list, return the node where the cycle begins. If there is no cycle, return null.
+ Follow up:
+ Can you solve it without using extra space?
+
+ Solution: ...
+*/
+
+/**
+* Definition for singly-linked list.
+* class ListNode {
+*     int val;
+*     ListNode next;
+*     ListNode(int x) {
+*         val = x;
+*         next = null;
+*     }
+* }
+*/
+public class Solution {
+    public ListNode detectCycle(ListNode head) {
+        if (head == null || head.next == null) return null;
+        ListNode slow = head, fast = head;
+        while (fast != null && fast.next != null) {
+            slow = slow.next;
+            fast = fast.next.next;
+            if (slow == fast) break;
+        }
+        if (fast == null || fast.next == null) return null;
+        slow = head;
+        while (slow != fast) {
+            slow = slow.next;
+            fast = fast.next;
+        }
+        return slow;
+    }
+}