Linked list in Java - 72: Rearrange a linked list in zig-zag manner

Source Code: thecodingsimpl...
Solution:
Take a boolean variable flag
Now, iterate from 1st node of list & check if element is less than next element
Same way, in next iteration check if it's greater than next element
In both condition, if condition matches, swap the values
At last, we'll have arranged value
Time Complexity: O(n) for iterating the n elements of linked list
Space Complexity: O(1) as we're taking only constant variables
Please check video for more info:
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[@ayeshaadhikari6123]

thank you sir! only your explanation was enough :)

[@CodingSimplified]

Thanks for your nice feedback. Keep Watching.

[@prasannapm3220]

thank you sir

[@rdxgaurav3483]

Node *zigZag(Node* head) { // your code goes here if(head == NULL || head->next == NULL){ return head; } Node* node = head; bool flag=true; while(node != NULL && node->next != NULL){ // if I am running without this node->next != NULL its giving segmentation fault , I don't know how the code in video is working but if we are at last node then the NULL->data will not be there and it will give runtime seg. error if(flag){ if(node->data > node->next->data){ swap(node->data, node->next->data); } flag = !flag; } else{ if(node->data < node->next->data){ swap(node->data , node->next->data); } flag = !flag; } node = node->next; } return head; }