Remove Duplicates from Unsorted List
Question
Write a removeDuplicates() function which takes a list and deletes
any duplicate nodes from the list. The list is not sorted.
For example if the linked list is 12->11->12->21->41->43->21,
then removeDuplicates() should convert the list to 12->11->21->41->43.
If temporary buffer is not allowed, how to solve it?
题解1 - 两重循环
Remove Duplicates 系列题，之前都是已排序链表，这个题为未排序链表。原题出自 CTCI 题2.1。
最容易想到的简单办法就是两重循环删除重复节点了，当前遍历节点作为第一重循环，当前节点的下一节点作为第二重循环。
Python
"""
Definition of ListNode
class ListNode(object):
def __init__(self, val, next=None):
self.val = val
self.next = next
"""
class Solution:
"""
@param head: A ListNode
@return: A ListNode
"""
def deleteDuplicates(self, head):
if head is None:
return None
curr = head
while curr is not None:
inner = curr
while inner.next is not None:
if inner.next.val == curr.val:
inner.next = inner.next.next
else:
inner = inner.next
curr = curr.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 *curr = head;
while (curr != NULL) {
ListNode *inner = curr;
while (inner->next != NULL) {
if (inner->next->val == curr->val) {
inner->next = inner->next->next;
} else {
inner = inner->next;
}
}
curr = curr->next;
}
return head;
}
};
Java
/**
* Definition for ListNode
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) {
* val = x;
* next = null;
* }
* }
*/
public class Solution {
/**
* @param ListNode head is the head of the linked list
* @return: ListNode head of linked list
*/
public static ListNode deleteDuplicates(ListNode head) {
if (head == null) return null;
ListNode curr = head;
while (curr != null) {
ListNode inner = curr;
while (inner.next != null) {
if (inner.next.val == curr.val) {
inner.next = inner.next.next;
} else {
inner = inner.next;
}
}
curr = curr.next;
}
return head;
}
}
源码分析
删除链表的操作一般判断node.next较为合适，循环时注意inner = inner.next和inner.next = inner.next.next的区别即可。
复杂度分析
两重循环，时间复杂度为 , 空间复杂度近似为 .
题解2 - 万能的 hashtable
使用辅助空间哈希表，节点值作为键，布尔值作为相应的值（是否为布尔值其实无所谓，关键是键）。
Python
"""
Definition of ListNode
class ListNode(object):
def __init__(self, val, next=None):
self.val = val
self.next = next
"""
class Solution:
"""
@param head: A ListNode
@return: A ListNode
"""
def deleteDuplicates(self, head):
if head is None:
return None
hash = {}
hash[head.val] = True
curr = head
while curr.next is not None:
if hash.has_key(curr.next.val):
curr.next = curr.next.next
else:
hash[curr.next.val] = True
curr = curr.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;
// C++ 11 use unordered_map
// unordered_map<int, bool> hash;
map<int, bool> hash;
hash[head->val] = true;
ListNode *curr = head;
while (curr->next != NULL) {
if (hash.find(curr->next->val) != hash.end()) {
ListNode *temp = curr->next;
curr->next = curr->next->next;
delete temp;
} else {
hash[curr->next->val] = true;
curr = curr->next;
}
}
return head;
}
};
Java
/**
* Definition for ListNode
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) {
* val = x;
* next = null;
* }
* }
*/
public class Solution {
/**
* @param ListNode head is the head of the linked list
* @return: ListNode head of linked list
*/
public static ListNode deleteDuplicates(ListNode head) {
if (head == null) return null;
ListNode curr = head;
HashMap<Integer, Boolean> hash = new HashMap<Integer, Boolean>();
hash.put(curr.val, true);
while (curr.next != null) {
if (hash.containsKey(curr.next.val)) {
curr.next = curr.next.next;
} else {
hash.put(curr.next.val, true);
curr = curr.next;
}
}
return head;
}
}
源码分析
删除链表中某个节点的经典模板在while循环中体现。
复杂度分析
遍历一次链表，时间复杂度为 , 使用了额外的哈希表，空间复杂度近似为 .
Reference
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