LeetCode刷题笔录Linked List Cycle II-优快云博客

博客围绕链表问题展开，给定一个链表，需找出循环起点，若不存在循环则返回特定值。采用slow和fast双指针法，先判定是否有循环，若有则让slow回到头节点，fast留在相遇点，二者以相同速度前进直至再次相遇，此相遇点即为循环起点。

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?

CTCI的原题。按照slow和fast双指针的方式先判定是否有cycle。如果slow==fast，即有cycle，那么跳出循环。让slow=head，fast还留在上一次的meet point。这时slow和fast以同一速度前进直到相遇。这个meet point就是cycle的起点。

/**
 * 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;
        ListNode fast = head;
        while(true){
            //no cycle
            if(fast.next == null || fast.next.next == null)
                return null;
            slow = slow.next;
            fast = fast.next.next;
            if(slow == fast)
                break;
        }
        //move slow to the head, while fast remain there
        //increment slow and fast one by one until they meet again
        //that meet point is the start of the cycle
        slow = head;
        while(slow != fast){
            slow = slow.next;
            fast = fast.next;
        }
        return slow;
        
    }
}