Leetcode 中Linked List Cycle 一类问题

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?

/**
 * 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){
            return false;
        }
        ListNode* slow = head;
        ListNode* fast = head;
        while (fast -> next != NULL && fast -> next -> next != NULL){    // 注意判断条件，第一次写的时候没有注意到要判断fast -> next是否为空
            slow = slow -> next;
            fast = fast -> next -> next;
            if (fast == slow){
                return true;
            }
        }
        return false;
    }
};

160. Intersection of Two Linked Lists

Write a program to find the node at which the intersection of two singly linked lists begins.

For example, the following two linked lists:

A:          a1 → a2
                   ↘
                     c1 → c2 → c3
                   ↗
B:     b1 → b2 → b3

begin to intersect at node c1.

Notes:

• If the two linked lists have no intersection at all, return null.
• The linked lists must retain their original structure after the function returns.
• You may assume there are no cycles anywhere in the entire linked structure.
• Your code should preferably run in O(n) time and use only O(1) memory.

两种方法：

1.  利用两个栈去做，但是这不符合题目O(1)空间复杂度的要求

2.  很tricky的一种方法。双指针解决：

Two pointer solution (O(n+m) running time, O(1) memory):

• Maintain two pointers pA and pB initialized at the head of A and B, respectively. Then let them both traverse through the lists, one node at a time.
• When pA reaches the end of a list, then redirect it to the head of B (yes, B, that's right.); similarly when pB reaches the end of a list, redirect it the head of A.
• If at any point pA meets pB, then pA/pB is the intersection node.
• To see why the above trick would work, consider the following two lists: A = {1,3,5,7,9,11} and B = {2,4,9,11}, which are intersected at node '9'. Since B.length (=4) < A.length (=6), pB would reach the end of the merged list first, because pB traverses exactly 2 nodes less than pA does. By redirecting pB to head A, and pA to head B, we now ask pB to travel exactly 2 more nodes than pA would. So in the second iteration, they are guaranteed to reach the intersection node at the same time.
• If two lists have intersection, then their last nodes must be the same one. So when pA/pB reaches the end of a list, record the last element of A/B respectively. If the two last elements are not the same one, then the two lists have no intersections.

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {
        if (headA == NULL || headB == NULL){
            return NULL;
        }
        ListNode* p1 = headA;
        ListNode* p2 = headB;

        while(p1 != p2){
            p1 = p1 == NULL ? headB: p1 -> next;   // 注意的一点是，这种条件判断也包含了当两个链表没有交点的情况，此时p1和p2都是NULL，会跳出循环，
            p2 = p2 == NULL ? headA: p2 -> next;
        }
        return p1;
    }
};

二次刷：思路是a和b都开始一步一步的走，然后如果a走到了链表的尾部就从b的头部从新开始走，同理，如果b走到了链表的尾部就从a的头部开始走，不过要注意的是要对到达过链表的尾部做标记，这样才能判断链表有没有交点。代码写的有点啰嗦，其实就是上边的代码的    简洁版。。。没有对比就没有提高

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {
        if (headA == NULL || headB == NULL){
            return NULL;
        }
        ListNode* pA = headA;
        ListNode* pB = headB;
        bool a = true;
        bool b = true;
        while (pA -> val != pB -> val){
            pA = pA -> next;
            pB = pB -> next;
            if (pA == NULL) {
                if (a == false){
                    return NULL;
                }
                pA = headB;
                a = false;

            }
            if (pB == NULL){
                if (b == false){
                    return NULL;
                }
                pB = headA;
                b = false;
            }

        }
        return pA;
    }
};

或者很正常的一种思路：先求出两个链表的长度，求出长度的差值，然后用两个指针，一个指针先走长度的差值的那些步数，然后两个指针开始一块走，碰到的话就是交点，到了末尾就是没有交点。

19. Remove Nth Node From End of List

Given a linked list, remove the nth node from the end of list and return its head.

For example,   Given linked list: 1->2->3->4->5, and n = 2.   After removing the second node from the end, the linked list becomes 1->2->3->5.

Note:
Given n will always be valid.
Try to do this in one pass.

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* removeNthFromEnd(ListNode* head, int n) {
        if (head == NULL){
            return NULL;
        }
        ListNode* newp = new ListNode();   // 要新建一个结点，来防止只有一个结点的情况出现
        newp -> next = head;

        ListNode* slow = newp;
        ListNode* fast = newp;

        for (int i = ; i <= n; i++){
            fast = fast -> next;
        }
        while (fast != NULL){
            slow = slow -> next;
            fast = fast -> next;
        }

        slow -> next = slow -> next -> next;
        return newp -> next;
    }
};

==========

142. Linked List Cycle II

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.

class Solution {
public:
    ListNode *detectCycle(ListNode *head) {
        if (!head) return NULL;
        ListNode *slow = head;
        ListNode *fast = head;
        bool HasCycle = false;
        while (fast->next != NULL && fast->next->next != NULL){
            slow = slow -> next;
            fast = fast -> next -> next;
            if (fast == slow){
                HasCycle = true;
                break;
            }
        }
        // 需要判断是否是有环的，若是有环，那么就
        if (HasCycle){
            // 若是有环，那么让fast 从头开始走，每次走一个结点。
            fast = head;
            while (slow != fast){
                fast = fast ->next;
                slow = slow ->next;
            }
            return fast;
        }
        return NULL;
    }
};