/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode() : val(0), next(nullptr) {}
* ListNode(int x) : val(x), next(nullptr) {}
* ListNode(int x, ListNode *next) : val(x), next(next) {}
* };
*/
class Solution {
public:
ListNode *deleteDuplicates(ListNode *head) {
if (head == nullptr || head->next == nullptr) {
return head;
}
// At least, we have two nodes in the linked list.
ListNode *dummy = new ListNode();
ListNode *prev = dummy, *left = head, *right = head;
dummy->next = head;
while (right != nullptr) {
if (left->val == right->val) {
right = right->next;
continue;
}
if (left->next == right) {
right = right->next;
left = left->next;
prev = prev->next;
continue;
}
deleteNodes(prev, left, right);
left = right;
}
if (left->next != right) {
deleteNodes(prev, left, right);
}
head = dummy->next;
delete dummy;
return head;
}
private:
void deleteNodes(ListNode *prev, ListNode *left, ListNode *right) {
prev->next = right;
while (left != right) {
ListNode *x = left;
left = left->next;
delete x;
}
}
};
head
.
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode() : val(0), next(nullptr) {}
* ListNode(int x) : val(x), next(nullptr) {}
* ListNode(int x, ListNode *next) : val(x), next(next) {}
* };
*/
class Solution {
public:
ListNode* deleteDuplicates(ListNode* head) {
ListNode *dummy = new ListNode(-200, head), *prev = dummy, *curr = head;
while (curr && curr->next) {
if (curr->val == curr->next->val) {
curr = curr->next;
continue;
}
if (prev->next == curr) {
prev = prev->next;
curr = curr->next;
} else {
curr = curr->next;
deleteDuplicates(prev, curr);
}
}
if (prev->next != curr) {
curr = curr->next;
deleteDuplicates(prev, curr);
}
head = dummy->next;
delete dummy;
return head;
}
private:
void deleteDuplicates(ListNode *left, ListNode *right) {
while (left->next != right) {
ListNode *x = left->next;
left->next = x->next;
delete x;
}
}
};
My Solution Solution 1: DFS (recursion) The Key Idea for Solving This Coding Question C++ Code /** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right;
Jun 6, 2025MyCircularQueueO(k)
Apr 20, 2025O(m)
Mar 4, 2025O(n)n is the length of nums.
Feb 19, 2025or
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