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：
1 // 2.1 Remove duplicates from a linked list 2 // inefficient without hashing space 3 #include <cstdio> 4 #include <unordered_set> 5 using namespace std; 6 7 struct ListNode { 8 int val; 9 struct ListNode *next; 10 ListNode(int x): val(x), next(nullptr) {}; 11 }; 12 13 class Solution { 14 public: 15 void removeDuplicates(ListNode *head) { 16 if (head == nullptr) { 17 return; 18 } 19 20 struct ListNode *ptr, *del, *tmp; 21 22 ptr = head; 23 while (ptr != nullptr && ptr->next != nullptr) { 24 del = ptr; 25 while (del->next != nullptr) { 26 if (ptr->val == del->next->val) { 27 tmp = del->next; 28 del->next = tmp->next; 29 delete tmp; 30 } else { 31 del = del->next; 32 } 33 } 34 ptr = ptr->next; 35 } 36 } 37 }; 38 39 int main() 40 { 41 int i; 42 int n; 43 int val; 44 struct ListNode *head, *ptr; 45 Solution sol; 46 47 while (scanf("%d", &n) == 1 && n > 0) { 48 // create a linked list 49 ptr = head = nullptr; 50 for (i = 0; i < n; ++i) { 51 scanf("%d", &val); 52 if (head == nullptr) { 53 head = ptr = new ListNode(val); 54 } else { 55 ptr->next = new ListNode(val); 56 ptr = ptr->next; 57 } 58 } 59 60 // remove duplicates from the list 61 sol.removeDuplicates(head); 62 63 // print the list 64 printf("%d", head->val); 65 ptr = head->next; 66 while (ptr != nullptr) { 67 printf("->%d", ptr->val); 68 ptr = ptr->next; 69 } 70 printf(" "); 71 72 // delete the list 73 while (head != nullptr) { 74 ptr = head->next; 75 delete head; 76 head = ptr; 77 } 78 } 79 80 return 0; 81 }
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：
1 // 2.1 Remove duplicates from a linked list 2 // efficient with hashing space 3 #include <cstdio> 4 #include <unordered_set> 5 using namespace std; 6 7 struct ListNode { 8 int val; 9 struct ListNode *next; 10 ListNode(int x): val(x), next(nullptr) {}; 11 }; 12 13 class Solution { 14 public: 15 void removeDuplicates(ListNode *head) { 16 if (head == nullptr) { 17 return; 18 } 19 20 unordered_set<int> us; 21 struct ListNode *ptr, *del; 22 23 us.insert(head->val); 24 ptr = head; 25 while (ptr->next != nullptr) { 26 if (us.find(ptr->next->val) != us.end()) { 27 // duplicate value 28 del = ptr->next; 29 ptr->next = del->next; 30 delete del; 31 } else { 32 ptr = ptr->next; 33 us.insert(ptr->val); 34 } 35 } 36 37 us.clear(); 38 } 39 }; 40 41 int main() 42 { 43 int i; 44 int n; 45 int val; 46 struct ListNode *head, *ptr; 47 Solution sol; 48 49 while (scanf("%d", &n) == 1 && n > 0) { 50 // create a linked list 51 ptr = head = nullptr; 52 for (i = 0; i < n; ++i) { 53 scanf("%d", &val); 54 if (head == nullptr) { 55 head = ptr = new ListNode(val); 56 } else { 57 ptr->next = new ListNode(val); 58 ptr = ptr->next; 59 } 60 } 61 62 // remove duplicates from the list 63 sol.removeDuplicates(head); 64 65 // print the list 66 printf("%d", head->val); 67 ptr = head->next; 68 while (ptr != nullptr) { 69 printf("->%d", ptr->val); 70 ptr = ptr->next; 71 } 72 printf(" "); 73 74 // delete the list 75 while (head != nullptr) { 76 ptr = head->next; 77 delete head; 78 head = ptr; 79 } 80 } 81 82 return 0; 83 }