《Cracking the Coding Interview》——第2章：链表——题目1

2014-03-18 02:16

题目：给定一个未排序的单链表，去除其中的重复元素。

解法1：不花额外空间，使用O(n^2)的比较方法来找出重复元素。

代码：

 // 2.1 Remove duplicates from a linked list
 // inefficient without hashing space
 #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:
     void removeDuplicates(ListNode *head) {
         if (head == nullptr) {
             return;
         }

         struct ListNode *ptr, *del, *tmp;

         ptr = head;
         while (ptr != nullptr && ptr->next != nullptr) {
             del = ptr;
             while (del->next != nullptr) {
                 if (ptr->val == del->next->val) {
                     tmp = del->next;
                     del->next = tmp->next;
                     delete tmp;
                 } else {
                     del = del->next;
                 }
             }
             ptr = ptr->next;
         }
     }
 };

 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;
             }
         }

         // remove duplicates from the list
         sol.removeDuplicates(head);

         // 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 ;
 }

解法2：使用额外空间的话，可以用unordered_set作为hash工具，进行重复元素的查找，效率更高。

代码：

 // 2.1 Remove duplicates from a linked list
 // efficient with hashing space
 #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:
     void removeDuplicates(ListNode *head) {
         if (head == nullptr) {
             return;
         }

         unordered_set<int> us;
         struct ListNode *ptr, *del;

         us.insert(head->val);
         ptr = head;
         while (ptr->next != nullptr) {
             if (us.find(ptr->next->val) != us.end()) {
                 // duplicate value
                 del = ptr->next;
                 ptr->next = del->next;
                 delete del;
             } else {
                 ptr = ptr->next;
                 us.insert(ptr->val);
             }
         }

         us.clear();
     }
 };

 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;
             }
         }

         // remove duplicates from the list
         sol.removeDuplicates(head);

         // 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 ;
 }