《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) == 1 && n > 0) {
        // create a linked list
        ptr = head = tail = nullptr;
        for (i = 0; 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 >= 1 && k <= n) {
            ptr = head;
            for (i = 1; 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
", first_node->val);
        } else {
            printf("no cycle
");
        }
        
        /*
        // print the list
        ptr = head;
        for (i = 0; i < n; ++i) {
            printf("%d->", ptr->val);
            ptr = ptr->next;
        }
        printf("
");
        */

        // delete the list
        for (i = 0; i < n; ++i) {
            ptr = head->next;
            delete head;
            head = ptr;
        }
    }
    
    return 0;
}

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) == 1 && n > 0) {
        // create a linked list
        ptr = head = tail = nullptr;
        for (i = 0; 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 >= 1 && k <= n) {
            ptr = head;
            for (i = 1; 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
", first_node->val);
        } else {
            printf("no cycle
");
        }
        
        /*
        // print the list
        ptr = head;
        for (i = 0; i < n; ++i) {
            printf("%d->", ptr->val);
            ptr = ptr->next;
        }
        printf("
");
        */

        // delete the list
        for (i = 0; i < n; ++i) {
            ptr = head->next;
            delete head;
            head = ptr;
        }
    }
    
    return 0;
}