Leetcode - LinedListCycleII

Given a linked list, return the node where the cycle begins. If there is no cycle, return null.

[分析] 这题可以说是快慢指针的经典题。首先判断是否存在环，而后找寻环的入口点。自己的naive思路是从快慢指针相交点处将链表分割成两个新链表，两个新链表的第一个交点即为环的入口点。更优化的思路是参考来的方法2，比较巧妙，但自己总是记不住，明明也理解了……优化法解析参考http://blog.csdn.net/kenden23/article/details/13871699

public class Solution {
    // method 2
    public ListNode detectCycle(ListNode head) {
        ListNode slow = head, fast = head;
        while (fast != null) {
            fast = fast.next;
            if (fast != null) {
                slow = slow.next;
                fast = fast.next;
                if (slow == fast)
                    break;
            } else {
                return null;
            }
        }
        if (fast == null)
            return null;
        slow = head;
        while (slow != fast) {
            slow = slow.next;
            fast = fast.next;
        }
        return slow;
    }
    // method 1
    public ListNode detectCycle1(ListNode head) {
        ListNode slow = head, fast = head;
        // step1: check if there is cycle
        while (fast != null) {
            fast = fast.next;
            if (fast != null) {
                slow = slow.next;
                fast = fast.next;
                if (slow == fast)
                    break;
            } else {
                return null;
            }
        }
        if (fast == null)
            return null;
        
        // step2: find the node which the cycle begins
        // split list at the meet node of slow and fast & search the first same node of the two list
        ListNode head2 = fast.next;
        fast.next = null;
        int len1 = 0, len2 = 0;
        ListNode p = head, q = head2;
        while (p != null) {
            p = p.next;
            len1++;
        }
        while (q != null) {
            q = q.next;
            len2++;
        }
        int diff = Math.abs(len1 - len2);
        p = head;
        q = head2;
        if (len1 > len2) {
            for (int i = 0; i < diff; i++)
                p = p.next;
        } else if (len1 < len2) {
            for (int i = 0; i < diff; i++)
                q = q.next;
        }
        while (p.val != q.val) {
            p = p.next;
            q = q.next;
        }
        fast.next = head2;
        return p;
    }
}