142. Linked List Cycle II

Problem:

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

To represent a cycle in the given linked list, we use an integer pos which represents the position (0-indexed) in the linked list where tail connects to. If pos is -1, then there is no cycle in the linked list.

Note: Do not modify the linked list.

Example 1:

Input: head = [3,2,0,-4], pos = 1
Output: tail connects to node index 1
Explanation: There is a cycle in the linked list, where tail connects to the second node.

Example 2:

Input: head = [1,2], pos = 0
Output: tail connects to node index 0
Explanation: There is a cycle in the linked list, where tail connects to the first node.

Example 3:

Input: head = [1], pos = -1
Output: no cycle
Explanation: There is no cycle in the linked list.

思路：
和141题类似的思路，也是设置fast和slow2个指针，fast每次移动2个节点，slow每次移动一个节点。设链表开始节点为start，环开始的节点为entry，fast和slow相遇的节点为meet。从start到entry距离为(x_1)，从entry到meet的距离为(x_2)，然后从meet沿环到entry的距离为(x_3)。则fast指针移动的距离为(x_1+x_2+x_3+x_2)，slow指针移动的距离为(x_1+x_2)，可以得出等式：(x_1+x_2+x_3+x_2 = 2(x_1+x_2))，可以推出(x_1=x_3)。也就是说，若设置一个新的节点to_entry从start开始移动，slow从meet开始移动，2者移动相同的距离就会在entry节点相遇，而此时移动的节点数也就是entry的下标，即找到了环开始的节点。

Solution:

ListNode *detectCycle(ListNode *head) {
    if (!head)  return NULL;
    
    ListNode *fast = head, *slow = head, *entry = head;
    
    while (fast->next && fast->next->next) {
        fast = fast->next->next;
        slow = slow->next;
        if (fast == slow) {
            while (slow != entry) {
                slow = slow->next;
                entry = entry->next;
            }
            return entry;
        }
    }
    return NULL;
}

性能：
Runtime: 12 ms  Memory Usage: 9.9 MB