The Solutions for LeetCode Problems

The solutions for LeetCode problems in C++ and Python.

LinkedList

No.21 Merge Two Sorted Lists

“Merge two sorted linked lists and return it as a new list. The new list should be made by splicing together the nodes of the first two lists.”

Time Complexity: O(m+n)

Space Complexity: O(1)

struct ListNode {
    int val;
    ListNode *next;
    ListNode(int x) : val(x), next(nullptr) {}
};

class Solution {
public:
    ListNode *mergeTwoLists(ListNode *l1, ListNode *l2) {
        ListNode *dummy = new ListNode(0);
        ListNode *tail = dummy;

        while (l1 && l2) {
            if (l1->val < l2->val) {
                tail->next = l1;
                l1 = l1->next;
            } else {
                tail->next = l2;
                l2 = l2->next;
            }
            tail = tail->next;
        }
        tail->next = l1 ? l1 : l2;

        return dummy->next;
    }
};

class ListNode:
    def __init__(self, x):
        self.val = x
        self.next = None

class Solution:
    def mergeTwoLists(self, l1: ListNode, l2: ListNode) -> ListNode:
        tail = dummy = ListNode(None)

        while l1 and l2:
            if l1.val < l2.val:
                tail.next = l1
                l1 = l1.next
            else:
                tail.next = l2
                l2 = l2.next

            tail = tail.next

        tail.next = l1 or l2

        return dummy.next

No.83 Remove Duplicates from Sorted List

“Given a sorted linked list, delete all duplicates such that each element appear
only once.”

struct ListNode {
    int val;
    ListNode *next;
    ListNode(int x) : val(x), next(nullptr) {}
};

class Solution {
public:
    ListNode *deleteDuplicates(ListNode *head) {
        ListNode *curr = head;
        while (curr && cur->next) {
            if (curr->val == curr->next->val) {
                curr->next = curr->next->next;
            } else {
                curr = curr->next;
            }
        }
        return head;
    }
};

class ListNode:
    def __init__(self, x):
        self.val = x
        self.next = None

class Solution:
    def deleteDuplicates(self, head: ListNode) -> ListNode:
        curr = head
        while curr and curr.next:
            if curr.val == curr.next.val:
                curr.next = curr.next.next
            else:
                curr = curr.next

        return head