Remove Duplicates from Sorted List
描述
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.
分析
无
解法1: 递归版
- Java
- C++
// Remove Duplicates from Sorted List
// 递归版，时间复杂度O(n)，空间复杂度O(n)
class Solution {
public ListNode deleteDuplicates(ListNode head) {
if (head == null || head.next == null) return head;
ListNode newHead = deleteDuplicates(head.next);
if(head.val == newHead.val) {
return newHead;
} else {
head.next=newHead;
}
return head;
}
};
// Remove Duplicates from Sorted List
// 递归版，时间复杂度O(n)，空间复杂度O(n)
class Solution {
public:
ListNode* deleteDuplicates(ListNode* head) {
if(head == nullptr || head->next == nullptr) return head;
ListNode* newHead = deleteDuplicates(head->next);
if(head->val == newHead->val) {
return newHead;
} else {
head->next=newHead;
}
return head;
}
};
解法2: 迭代版
- Java
- C++
// Remove Duplicates from Sorted List
// 迭代版，时间复杂度O(n)，空间复杂度O(1)
class Solution {
public ListNode deleteDuplicates(ListNode head) {
if (head == null) return null;
for (ListNode slow = head, fast = head.next; fast != null; fast = slow.next) {
if (slow.val == fast.val) {
slow.next = fast.next;
} else {
slow = fast;
}
}
return head;
}
};
// Remove Duplicates from Sorted List
// 迭代版，时间复杂度O(n)，空间复杂度O(1)
class Solution {
public:
ListNode *deleteDuplicates(ListNode *head) {
if (head == nullptr) return nullptr;
for (ListNode *slow = head, *fast = head->next; fast != nullptr; fast = slow->next) {
if (slow->val == fast->val) {
slow->next = fast->next;
delete fast;
} else {
slow = fast;
}
}
return head;
}
};