Remove Duplicates from Sorted List
Source
Given a sorted linked list, delete all duplicates such that each element appear only once. Example Given 1->1->2, return 1->2. Given 1->1->2->3->3, return 1->2->3.
题解
遍历之,遇到当前节点和下一节点的值相同时,删除下一节点,并将当前节点next
值指向下一个节点的next
, 当前节点首先保持不变,直到相邻节点的值不等时才移动到下一节点。
C++
/** * Definition of ListNode * class ListNode { * public: * int val; * ListNode *next; * ListNode(int val) { * this->val = val; * this->next = NULL; * } * } */ class Solution { public: /** * @param head: The first node of linked list. * @return: head node */ ListNode *deleteDuplicates(ListNode *head) { if (head == NULL) { return NULL; } ListNode *node = head; while (node->next != NULL) { if (node->val == node->next->val) { ListNode *temp = node->next; node->next = node->next->next; delete temp; } else { node = node->next; } } return head; } };
Java
/** * Definition for singly-linked list. * public class ListNode { * int val; * ListNode next; * ListNode(int x) { val = x; } * } */ public class Solution { public ListNode deleteDuplicates(ListNode head) { if (head == null) return null; ListNode node = head; while (node.next != null) { if (node.val == node.next.val) { node.next = node.next.next; } else { node = node.next; } } return head; } }
源码分析
- 首先进行异常处理,判断head是否为NULL
- 遍历链表,
node->val == node->next->val
时,保存node->next
,便于后面释放内存(非C/C++无需手动管理内存) - 不相等时移动当前节点至下一节点,注意这个步骤必须包含在
else
中,否则逻辑较为复杂
while
循环处也可使用node != null && node->next != null
, 这样就不用单独判断head
是否为空了,但是这样会降低遍历的效率,因为需要判断两处。
复杂度分析
遍历链表一次,时间复杂度为 O(n), 使用了一个中间变量进行遍历,空间复杂度为 O(1).