Leetcode NO.83 Remove Duplicates from Sorted List
2015-02-05 12:17
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本题的题目要求如下:
Given a sorted linked list, delete all duplicates such that each element appear only once.
For example,
Given
Given
这是我第二次做这道简单题,发现这次写的代码比上次要简洁很多,看来,刷leetcode的这两个月还是有一定的长进的。。
举一个例子说明本题的基本算法:
1->1->1->1->2->3
取一个指针ptr1指向第一个1,另外一个ptr2指向这个指针的next,如果跟ptr2指向的和ptr1指针指向的元素相等,则将ptr2继续向后移动一位,直到不一样位置,比如,此时ptr2指向2,这是讲ptr1->next指向现在的ptr2,之后重复这种操作。。
详细代码如下:
上一个版本应该是有内存泄漏的,现在写了一个应该避免了内存泄漏的版本。。如果大家觉得还是不对,希望不吝赐教:
Given a sorted linked list, delete all duplicates such that each element appear only once.
For example,
Given
1->1->2, return
1->2.
Given
1->1->2->3->3, return
1->2->3.
这是我第二次做这道简单题,发现这次写的代码比上次要简洁很多,看来,刷leetcode的这两个月还是有一定的长进的。。
举一个例子说明本题的基本算法:
1->1->1->1->2->3
取一个指针ptr1指向第一个1,另外一个ptr2指向这个指针的next,如果跟ptr2指向的和ptr1指针指向的元素相等,则将ptr2继续向后移动一位,直到不一样位置,比如,此时ptr2指向2,这是讲ptr1->next指向现在的ptr2,之后重复这种操作。。
详细代码如下:
<pre name="code" class="cpp">/** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} * }; */ class Solution { public: ListNode* deleteDuplicates(ListNode* head) { if (head == nullptr) { return head; } ListNode* virtual_head = new ListNode(-1); virtual_head->next = head; int tmp = head->val; ListNode* prev = head; head = head->next; while (head != nullptr) { if (tmp != head->val) { prev->next = head; prev = head; tmp = prev->val; } head = head->next; } prev->next = head; return virtual_head->next; } };
上一个版本应该是有内存泄漏的,现在写了一个应该避免了内存泄漏的版本。。如果大家觉得还是不对,希望不吝赐教:
/** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} * }; */ class Solution { public: ListNode* deleteDuplicates(ListNode* head) { ListNode* cur = head; while (cur != nullptr) { ListNode* tmp = cur->next; while (tmp != nullptr and cur->val == tmp->val) { ListNode* del = tmp; tmp = tmp->next; cur->next = tmp; delete del; } cur = cur->next; } return head; } };
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