# Remove Duplicates from Sorted List ### Source - leetcode: [Remove Duplicates from Sorted List | LeetCode OJ](https://leetcode.com/problems/remove-duplicates-from-sorted-list/) - lintcode: [(112) Remove Duplicates from Sorted List](http://www.lintcode.com/en/problem/remove-duplicates-from-sorted-list/) ~~~ 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。 ### Python ~~~ # Definition for singly-linked list. # class ListNode: # def __init__(self, x): # self.val = x # self.next = None class Solution: # @param {ListNode} head # @return {ListNode} def deleteDuplicates(self, head): if head is None: return None node = head while node.next is not None: if node.val == node.next.val: node.next = node.next.next else: node = node.next return head ~~~ ### 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 1. 首先进行异常处理，判断head是否为NULL 1. 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) 1. 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)O(n)O(n), 使用了一个中间变量进行遍历，空间复杂度为 O(1)O(1)O(1). ### Reference - [Remove Duplicates from Sorted List 参考程序 | 九章](http://www.jiuzhang.com/solutions/remove-duplicates-from-sorted-list/)