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May 16, 2020 17:46
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Day 15:LeetCode 30 Day May challenges
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| /** | |
| * 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* oddEvenList(ListNode* head) { | |
| if(head == NULL) | |
| return head; | |
| ListNode* odd = head; | |
| ListNode* evennode = head->next; | |
| ListNode* even = evennode; | |
| while(even!=NULL && even->next != NULL){ | |
| odd->next = (odd->next)->next; | |
| even->next = (even->next)->next; | |
| odd = odd->next; | |
| even = even->next; | |
| } | |
| //combining two lists | |
| odd->next = evennode; | |
| return head; | |
| } | |
| }; | |
| /* | |
| ListNode* GetnewNode(int data){ | |
| ListNode* temp = new ListNode(data); | |
| return temp; | |
| } | |
| ListNode* Insert(ListNode* root,int data){ | |
| //create new node | |
| ListNode* temp = GetnewNode(data); | |
| if(root == NULL){ | |
| root = temp; | |
| return root; | |
| } | |
| //Inserting at tail | |
| ListNode* temp_1 = root; | |
| while(temp_1->next != NULL){ | |
| temp_1 = temp_1->next; | |
| } | |
| temp_1->next = temp; | |
| return root; | |
| } | |
| ListNode* oddEvenList(ListNode* head) { | |
| //create odd group linkedlist(1->3->5->....) | |
| //create even group linkedlist(2->4->6->8...) | |
| //concatenate them | |
| if(head == NULL) | |
| return head; | |
| ListNode* temp = head; | |
| //root node must be pointed to null | |
| ListNode* root_1 = NULL; | |
| ListNode* root_2 = NULL; | |
| int turn = 1; | |
| while(temp != NULL){ | |
| if(turn%2 == 0){ | |
| //even nodes list | |
| root_2 = Insert(root_2,temp->val); | |
| } | |
| else{ | |
| //odd nodes list | |
| root_1 = Insert(root_1,temp->val); | |
| } | |
| temp = temp->next; | |
| turn+=1; | |
| } | |
| //joining two lists | |
| ListNode* temp_1 = root_1; | |
| while(temp_1->next != NULL){ | |
| temp_1 = temp_1->next; | |
| } | |
| temp_1->next = root_2; | |
| return root_1; | |
| } | |
| */ |
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| Given a singly linked list, group all odd nodes together followed by the even nodes. Please note here we are talking about the node number and not the value in the nodes. | |
| You should try to do it in place. The program should run in O(1) space complexity and O(nodes) time complexity. | |
| Example 1: | |
| Input: 1->2->3->4->5->NULL | |
| Output: 1->3->5->2->4->NULL | |
| Example 2: | |
| Input: 2->1->3->5->6->4->7->NULL | |
| Output: 2->3->6->7->1->5->4->NULL | |
| Note: | |
| The relative order inside both the even and odd groups should remain as it was in the input. | |
| The first node is considered odd, the second node even and so on ... |
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