鏈表中與環相關的問題

141. Linked List Cycle

Given a linked list, determine if it has a cycle in it.

Follow up:
Can you solve it without using extra space?

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題目鏈接

利用一個快指針，一個慢指針。如果它們能相遇，則存在環。

/**
 * Language: cpp ( 9ms )
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    bool hasCycle(ListNode *head) {

        if(head==NULL||head->next==NULL){
            return false;
        }
        ListNode* fast=head,*slow=head->next;
        while(fast!=slow){
            if(slow->next==NULL||slow->next->next==NULL){
                return false;
            }else{
                fast=fast->next;
                slow=slow->next->next;
            }

        }
        return true;
    }
};

142. Linked List Cycle II

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?

題目鏈接

判斷環的入口點 ：相遇點到環入口的距離等於頭節點到環入口的距離。

class Solution {
public:
    ListNode *detectCycle(ListNode *head) {
        if(head==NULL||head->next==NULL){
            return NULL;
        }
        ListNode *slow=head,*fast=head;
        while(fast!=NULL&&fast->next!=NULL){
            fast=fast->next->next;
            slow=slow->next;
            if(fast==slow){
                break;
            }
        }
        if(fast!=slow){
            return NULL;
        }
        slow=head;
        while(fast!=slow){
            fast=fast->next;
            slow=slow->next;
        }
        return fast;
    }
};

計算環的長度
如何判斷兩個鏈表（不帶環）是否相交
資料參考