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Create List Cycle .java

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	`1`	`+/*`
	`2`	`+Given a linked list, return the node where the cycle begins. If there is no cycle, return null.`
	`3`	`+`
	`4`	`+Try solving it using constant additional space.`
	`5`	`+`
	`6`	`+Example :`
	`7`	`+`
	`8`	`+Input :`
	`9`	`+`
	`10`	`+ ______`
	`11`	`+ \| \|`
	`12`	`+ \/ \|`
	`13`	`+ 1 -> 2 -> 3 -> 4`
	`14`	`+`
	`15`	`+Return the node corresponding to node 3.`
	`16`	`+*/`
	`17`	`+`
	`18`	`+/**`
	`19`	`+ * Definition for singly-linked list.`
	`20`	`+ * class ListNode {`
	`21`	`+ * public int val;`
	`22`	`+ * public ListNode next;`
	`23`	`+ * ListNode(int x) { val = x; next = null; }`
	`24`	`+ * }`
	`25`	`+ */`
	`26`	`+public class Solution {`
	`27`	`+ public ListNode detectCycle(ListNode a) {`
	`28`	`+ ListNode slow = a;`
	`29`	`+ ListNode fast = a;`
	`30`	`+`
	`31`	`+ while (fast != null && fast.next != null) {`
	`32`	`+ slow = slow.next;`
	`33`	`+ fast = fast.next.next;`
	`34`	`+`
	`35`	`+ if (slow == fast) {`
	`36`	`+ break;`
	`37`	`+ }`
	`38`	`+ }`
	`39`	`+`
	`40`	`+ if (fast == null \|\| fast.next == null) return null;`
	`41`	`+`
	`42`	`+ slow = a;`
	`43`	`+ while (slow != fast) {`
	`44`	`+ slow = slow.next;`
	`45`	`+ fast = fast.next;`
	`46`	`+ }`
	`47`	`+`
	`48`	`+ return slow;`
	`49`	`+ }`
	`50`	`+}`

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