141. Linked List Cycle
题目：Given a linked list, determine if it has a cycle in it.
Follow up: Can you solve it without using extra space?
分析：要确定一个链表中是否有环，设置两个指针，一个fast，一个慢slow，如果fast和slow相遇则证明有环。
代码：
/** * Definition for singly-linked list. * class ListNode { * int val; * ListNode next; * ListNode(int x) { * val = x; * next = null; * } * } */ public class Solution { public boolean hasCycle(ListNode head) { ListNode fast = head; ListNode slow = head; while(fast!=null&&fast.next!=null){ fast = fast.next.next; slow = slow.next; if (fast==slow) return true; } return false; } }
142. Linked List Cycle II题目：
Given a linked list, return the node where the cycle begins. If there is no cycle, return null.
Note: Do not modify the linked list.
Follow up: Can you solve it without using extra space?
分析：
当fast若与slow相遇时，slow肯定没有走遍历完链表，而fast已经在环内循环了n圈(1<=n)。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，入口环与相遇点距离为x，起点到环入口点的距离为a。 a + x = nr a + x = (n – 1)r +r = (n-1)r + L - a a = (n-1)r + (L – a – x)
(L – a – x)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)循环内环+相遇点到环入口点，于是我们从链表头、与相遇点分别设一个指针，每次各走一步，两个指针必定相遇，且相遇第一点为环入口点。
代码：
/** * 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) { ListNode fast = head; ListNode slow = head; while(fast!=null&&fast.next!=null){ fast = fast.next.next; slow = slow.next; if (fast==slow) break; } if((fast==null)||(fast.next==null)) return null; slow = head; while(slow != fast){ slow = slow.next; fast = fast.next; } return slow; } }
