《Cracking the Coding Interview》——第2章：链表——题目7

2014-03-18 02:57

题目：检查链表是否是回文的，即是否中心对称。

解法：我的做法是将链表从中间对半拆成两条，然后把后半条反转，再与前半条对比。对比完了再将后半条反转了拼回去。这样不涉及额外的空间，比反转整条链表然后比较要来的好。如果你反转了整条链表又不用额外空间，接下来跟谁比去呢？

代码：

 // 2.7 To Check the given linked list is palindrome or not?
 #include <cstdio>
 #include <unordered_set>
 using namespace std;

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

 class Solution {
 public:
     bool isPalindromeList(ListNode *head) {
         if (head == nullptr) {
             return false;
         }
         if (head->next == nullptr) {
             return true;
         }

         ListNode *t1, *run, *h2;

         t1 = run = head;
         while (run->next != nullptr && run->next->next != nullptr) {
             t1 = t1->next;
             run = run->next->next;
         }
         h2 = t1->next;
         t1->next = nullptr;
         h2 = reverseList(h2);

         ListNode *p1, *p2;
         p1 = head;
         p2 = h2;
         while (p2 != nullptr) {
             if (p1->val != p2->val) {
                 break;
             } else {
                 p1 = p1->next;
                 p2 = p2->next;
             }
         }
         bool res = (p2 == nullptr);

         h2 = reverseList(h2);
         t1->next = h2;

         return res;
     }
 private:
     ListNode* reverseList(ListNode* head) {
         if (head == nullptr) {
             return head;
         }

         ListNode* new_head = nullptr;
         ListNode* ptr;

         while (head != nullptr) {
             if (new_head == nullptr) {
                 new_head = head;
                 head = head->next;
                 new_head->next = nullptr;
             } else {
                 ptr = head->next;
                 head->next = new_head;
                 new_head = head;
                 head = ptr;
             }
         }

         return new_head;
     }
 };

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

     while (scanf("%d", &n) ==  && n > ) {
         // create a linked list
         ptr = head = 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;
             }
         }

         // check if the list is a palindrome
         if (sol.isPalindromeList(head)) {
             printf("It's a palindrome.\n");
         } else {
             printf("It's not a palindrome.\n");
         }

         // print the list
         printf("%d", head->val);
         ptr = head->next;
         while (ptr != nullptr) {
             printf("->%d", ptr->val);
             ptr = ptr->next;
         }
         printf("\n");

         // delete the list
         while (head != nullptr) {
             ptr = head->next;
             delete head;
             head = ptr;
         }
     }

     return ;
 }