leetcode 141 142 Linked List Cycle I/II

使用快慢指针，如果有环快慢指针一定会相遇

class Solution {
public:
bool hasCycle(ListNode *head) {
ListNode *fast, *slow;
fast = slow = head;
while(slow && fast) {
slow = slow -> next;
if(fast -> next)
fast = fast -> next -> next; //bug处，如果没有if条件跳bug：fast幅值给了空值
else
return false;
if(fast == slow)
return true;
}
return false;
}
};

class Solution {//不是最优解 用了O（n）的空间
public:
ListNode *detectCycle(ListNode *head) {
ListNode *move_forward = head;
int count = 0;
set<ListNode *> iset;//bug处 最初把set类型设为struct listnode 不合法因为set元素必须为可比较类型
while(move_forward) {
iset.insert(move_forward);
++count;
if(iset.size() != count)
return move_forward;
move_forward = move_forward -> next;
}
return nullptr;
}
};

Alogrithm Description:

Step 1: Determine whether there is a cycle

1.1) Using a slow pointer that move forward 1 step each time

1.2) Using a fast pointer that move forward 2 steps each time

1.3) If the slow pointer and fast pointer both point to the same location after several moving steps, there is a cycle;

1.4) Otherwise, if (fast->next == NULL || fast->next->next == NULL), there has no cycle.

Step 2: If there is a cycle, return the entry location of the cycle

2.1) L1 is defined as the distance between the head point and entry point

2.2) L2 is defined as the distance between the entry point and the meeting point

2.3) C is defined as the length of the cycle

2.4) n is defined as the travel times of the fast pointer around the cycle When the first encounter of the slow pointer and the fast pointer

According to the definition of L1, L2 and C, we can obtain:

• the total distance of the slow pointer traveled when encounter is L1 + L2

• the total distance of the fast pointer traveled when encounter is L1 + L2 + n * C

• Because the total distance the fast pointer traveled is twice as the slow pointer, Thus:

• 2 * (L1+L2) = L1 + L2 + n * C => L1 + L2 = n * C => L1 = (n - 1) C + (C - L2)*

It can be concluded that the distance between the head location and entry location is equal to the distance between the meeting location and the entry location along the direction of forward movement.

So, when the slow pointer and the fast pointer encounter in the cycle, we can define a pointer "entry" that point to the head, this "entry" pointer moves one step each time so as the slow pointer. When this "entry" pointer and the slow pointer both point to the same location, this location is the node where the cycle begins.

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Here is the code:

ListNode *detectCycle(ListNode *head) {
if (head == NULL || head->next == NULL)
return NULL;

ListNode *slow = head;
ListNode *fast = head;
ListNode *entry = head;

while (fast->next && fast->next->next) {
slow = slow->next;
fast = fast->next->next;
if (slow == fast) { // there is a cycle
while(slow != entry) { // found the entry location
slow = slow->next;
entry = entry->next;
}
return entry;
}
}
return NULL; // there has no cycle
}