Linked List Cycle II
描述
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?
分析
当fast与slow相遇时,slow肯定没有遍历完链表,而fast已经在环内循环了n
圈()。假设slow走了s
步,则fast走了2s
步(fast步数还等于s
加上在环上多转的n
圈),设环长为r
,则:
2s = s + nr
s = nr
设整个链表长L
,环入口点与相遇点距离为a
,起点到环入口点的距离为x
,则
x + a = nr = (n – 1)r +r = (n-1)r + L - x
x = (n-1)r + (L – x – a)
L – x – a
为相遇点到环入口点的距离,由此可知,从链表头到环入口点等于n-1
圈内环+相遇点到环入口点,于是我们可以从head
开始另设一个指针slow2
,两个慢指针每次前进一步,它俩一定会在环入口点相遇。
代码
// Linked List Cycle II
// 时间复杂度O(n),空间复杂度O(1)
class Solution {
public:
ListNode *detectCycle(ListNode *head) {
ListNode *slow = head, *fast = head;
while (fast && fast->next) {
slow = slow->next;
fast = fast->next->next;
if (slow == fast) {
ListNode *slow2 = head;
while (slow2 != slow) {
slow2 = slow2->next;
slow = slow->next;
}
return slow2;
}
}
return nullptr;
}
};