《Cracking the Coding Interview》——第2章：链表——题目6

2014-03-18 02:41

题目：给定一个带有环的单链表，找出环的入口节点。

解法1：用hash来检测重复节点肯定是容易想而且效率也高的好办法。

代码：

 // 2.6 You have a circular Linked List: a->b->c->d->e->c. Find where the cycle starts.
 #include <cstdio>
 #include <unordered_set>
 using namespace std;

 struct ListNode {
     int val;
     ListNode *next;
     ListNode(int x): val(x), next(nullptr) {};
 };

 class Solution {
 public:
     ListNode* firstFirstNodeInCycle(ListNode *head) {
         if (head == nullptr) {
             return head;
         }

         // hash the pointers.
         unordered_set<ListNode *> us;
         ListNode *ptr;

         ptr = head;
         while (ptr != nullptr) {
             if (us.find(ptr) != us.end()) {
                 // the first node of the cycle is found.
                 return ptr;
             } else {
                 us.insert(ptr);
                 ptr = ptr->next;
             }
         }

         // the list has no cycle.
         return nullptr;
     }
 };

 int main()
 {
     int i;
     int n, k;
     int val;
     struct ListNode *head, *tail, *ptr;
     Solution sol;

     while (scanf("%d", &n) ==  && n > ) {
         // create a linked list
         ptr = head = tail = nullptr;
         for (i = ; i < n; ++i) {
             scanf("%d", &val);
             if (head == nullptr) {
                 head = ptr = new ListNode(val);
             } else {
                 ptr->next = new ListNode(val);
                 ptr = ptr->next;
             }
         }
         tail = ptr;

         // create a cycle in the list
         scanf("%d", &k);
         if (k >=  && k <= n) {
             ptr = head;
             for (i = ; i < k; ++i) {
                 ptr = ptr->next;
             }
             tail->next = ptr;
         }

         // find the first node in the cycle.
         ListNode *first_node = sol.firstFirstNodeInCycle(head);
         if (first_node != nullptr) {
             printf("%d\n", first_node->val);
         } else {
             printf("no cycle\n");
         }

         /*
         // print the list
         ptr = head;
         for (i = 0; i < n; ++i) {
             printf("%d->", ptr->val);
             ptr = ptr->next;
         }
         printf("\n");
         */

         // delete the list
         for (i = ; i < n; ++i) {
             ptr = head->next;
             delete head;
             head = ptr;
         }
     }

     return ;
 }

解法2：如果你非不让我用hash，就只有观察一下几个特殊节点了。既然这个环有“第一个”点，那肯定也有“最后一个”点。我特地打了引号，是因为最后一个点恰好指向第一个。如果你从链表的头开始向后遍历，必然是先到达“第一个”，然后才能到达“最后一个的”。只有对于“最后一个”点，才会出现先到达ptr->next，后到达ptr的情况，所以如果存在符合这个情况的节点ptr，ptr->next就是我们要找的环的入口。这种方法效率不高而且不容易想，除了面试之外基本没有别的用处了。但为了面试淘汰一些人，这点用处也很重要了。

代码：

 // 2.6 You have a circular Linked List: a->b->c->d->e->c. Find where the cycle starts.
 #include <cstdio>
 #include <unordered_set>
 using namespace std;

 struct ListNode {
     int val;
     ListNode *next;
     ListNode(int x): val(x), next(nullptr) {};
 };

 class Solution {
 public:
     ListNode* firstFirstNodeInCycle(ListNode *head) {
         if (head == nullptr) {
             return head;
         }

         ListNode *p1, *p2;

         p1 = head;
         while (p1 != nullptr) {
             p2 = head;
             while (p2 != p1->next && p2 != nullptr) {
                 if (p2 == p1) {
                     break;
                 } else {
                     p2 = p2->next;
                 }
             }
             if (p2 == p1->next) {
                 return p2;
             } else {
                 p1 = p1->next;
             }
         }

         return nullptr;
     }
 };

 int main()
 {
     int i;
     int n, k;
     int val;
     struct ListNode *head, *tail, *ptr;
     Solution sol;

     while (scanf("%d", &n) ==  && n > ) {
         // create a linked list
         ptr = head = tail = nullptr;
         for (i = ; i < n; ++i) {
             scanf("%d", &val);
             if (head == nullptr) {
                 head = ptr = new ListNode(val);
             } else {
                 ptr->next = new ListNode(val);
                 ptr = ptr->next;
             }
         }
         tail = ptr;

         // create a cycle in the list
         scanf("%d", &k);
         if (k >=  && k <= n) {
             ptr = head;
             for (i = ; i < k; ++i) {
                 ptr = ptr->next;
             }
             tail->next = ptr;
         }

         // find the first node in the cycle.
         ListNode *first_node = sol.firstFirstNodeInCycle(head);
         if (first_node != nullptr) {
             printf("%d\n", first_node->val);
         } else {
             printf("no cycle\n");
         }

         /*
         // print the list
         ptr = head;
         for (i = 0; i < n; ++i) {
             printf("%d->", ptr->val);
             ptr = ptr->next;
         }
         printf("\n");
         */

         // delete the list
         for (i = ; i < n; ++i) {
             ptr = head->next;
             delete head;
             head = ptr;
         }
     }

     return ;
 }