[LeetCode] 142. Linked List Cycle II 链表中的环 II

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?

141. Linked List Cycle 的拓展，这题要返回环开始的节点，如果没有环返回null。

解法：双指针，还是用快慢两个指针，相遇时记下节点。参考：willduan的博客

Java：

public class Solution {
    public ListNode detectCycle( ListNode head ) {
        if( head == null || head.next == null ){
            return null;
        }
        // 快指针fp和慢指针sp，
        ListNode fp = head, sp = head;
        while( fp != null && fp.next != null){
            sp = sp.next;
            fp = fp.next.next;
            //此处应该用fp == sp ，而不能用fp.equals(sp) 因为链表为1 2 的时候容易
            //抛出异常
            if( fp == sp ){  //说明有环
                break;
            }
        }
        //System.out.println( fp.val + "   "+ sp.val );
        if( fp == null || fp.next == null ){
            return null;
        }
        //说明有环，求环的起始节点
        sp = head;
        while( fp != sp ){
            sp = sp.next;
            fp = fp.next;
        }
        return sp;
    }
} 　　

Python：

class ListNode:
    def __init__(self, x):
        self.val = x
        self.next = None

    def __str__(self):
        if self:
            return "{}".format(self.val)
        else:
            return None

class Solution:
    # @param head, a ListNode
    # @return a list node
    def detectCycle(self, head):
        fast, slow = head, head
        while fast and fast.next:
            fast, slow = fast.next.next, slow.next
            if fast is slow:
                fast = head
                while fast is not slow:
                    fast, slow = fast.next, slow.next
                return fast
        return None　　

C++：

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) break;
        }
        if (!fast || !fast->next) return NULL;
        slow = head;
        while (slow != fast) {
            slow = slow->next;
            fast = fast->next;
        }
        return fast;
    }
};

　　

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