yujinred Here is my solution in C++. It's pretty simple.
if (headA == NULL && headB == NULL) { return 1; } else if (headA == NULL || headB == NULL) { return 0; } if (headA->data == headB->data) { return CompareLists(headA->next, headB->next); } else { return 0; }
neo_3 Improved version
if (headA == NULL && headB == NULL) { return 1; } else if (headA == NULL || headB == NULL) { return 0; } return (headA->data == headB->data) && CompareLists(headA->next, headB->next); }
sushantsagar29 My solution. Simpler.
int CompareLists(Node *headA, Node* headB) { while(headA!=NULL && headB!=NULL && headA->data==headB->data)headA=headA->next,headB=headB->next; return headA==headB; }
destiner Kudos for no recursion!
rajabca2011 We should not check the address of headA & headB and it is wrong comparison.
sushantsagar29 Reason behind not doing that?
asperatology The test cases are flawed. You're not supposed to compare the addresses because who knows if the linked lists are placed on different heaps of different pages in the memory.
GauravEic Hmm, we need to check for headA.data vs headB.data
erenyaeger2017 Actually we should, @sushantsagar29's logic is correct. This is because at the point of return, both pointers should be null if they are identical. If they aren't NULL, means we did not reach the end of both linked list. They are identical only if we reach the end of the two linked list
vishnuPrasanna wow..kidu
logan789 wow
cayhan it is not solution, unfortunately.
themast3r int CompareLists(Node *headA, Node* headB) { while(headA && headB) { if(headA -> data != headB -> data) return 0; headA = headA->next; headB = headB -> next; } return(!(headA || headB)); }
shantanusalil what does "return(!(headA || headB))" mean
mkmarathe1 a check that headA and headB both equal. They being equal would pass the test cases, but an explicit check with NULL would be better.
CoDe_GeEk_ upvoted for return(!(headA || headB))
positivedeist_07 This is a very clean code :) congrats!! But will it work for all the cases? Coz the problem has only 2 test cases. I'm just curious, that's all :) Why do use return headA==headB at the end? i guess it works like an if condition?? pls correct me if i'm wrong. Overall, brilliant thinking!
alaniane I would not say that your solution is simpler to understand. The while statement is overly complex. Better spacing between the operators would help with readability.
"headA != NULL" instead of "headA!=NULL"
At a cursory glance I thought you were accidentally assigning NULL to the pointer until I saw the ! operator.
while(...) headA=headA->next, headB=headB-next;
Although you can use the comma operator to force the statement into one line, a clearer implementation would be:
while(...) { headA=headA->next; headB=headB->next; }
However, what I like about your solution is that it does not use recursion.
prateekkk [deleted]sanjayrockstar26 please explAIN THE code to me
Mr_BrainTurner wow
ujjwalsaxena My solution in Java. Not sure whether optimal or not
int CompareLists(Node headA, Node headB) { if(headA!=null && headB!=null) { if(headA.data==headB.data) return CompareLists(headA.next,headB.next); else return 0; } else if(headA==null && headB==null) // both null at the same time return 1; else return 0; // if any one is null }
14p350 return CompareLists(headA.next,headB.next); what does this return?please explain
ujjwalsaxena its a recursive call, CompareLists is the name of the method i created in the first line of code, and I am calling it again passing values the next node of both headA and headB
riteshsingh0097 pretty one.
Tara_123 Improved version
int CompareLists(Node *headA, Node* headB){ return((headA==NULL && headB==NULL)?1:((headA==NULL || headB==NULL)?0:(headA->data==headB->data)&&(CompareLists(headA->next,headB->next)))); }
shivamfromkanpur I think your solution is complex then 'yujinred' because in your solution we have to go untill null .what ever that data equal or no.
Abirkhan88821 ohhh its great
stark007_sc A pretty nice one
kameshshahit I feel ashamed after seeing this code.such amazing and simple code .i have written it like kindergarden kid. God help me!!
ishabandi You are not alone! Here is mine:
def CompareLists(headA, headB): if headA is None or headB is None: return 0 else: nodeA = headA nodeB = headB flag = 0 while nodeA is not None and nodeB is not None: if nodeA.data != nodeB.data: flag = 2 nodeA= nodeA.next nodeB=nodeB.next if flag == 2: return 0 else: if nodeA is None and nodeB is not None: return 0 elif nodeB is None and nodeA is not None: return 0 else: return 1
riwiem I like to solve it in python too:
def CompareLists(headA, headB): while True: # both nodes are None if (not headA) and (not headB): return 1 # only one node is None or node data differs if (bool(headA) != bool(headB)) or (headA.data != headB.data): return 0 headA = headA.next headB = headB.next
First check if both nodes are None, which implies that all previous nodes (if any) are identical. If that's not the case, check if (only) one list is at it's tail or if the data of the current nodes differs. If neither is true, we can move both current nodes forward and repeat the loop.
jae_duk_seo this a nice non recursive way
surendraps007 how you deal with list that contain diffrent lenght of headA and headB
shreya1sisodia [deleted]shreya1sisodia [deleted]shreya1sisodia tried to make it shorter
int CompareLists(Node headA, Node headB)
{ if(headA==NULL &&headB==NULL)
return 1; if((headA==NULL||headB==NULL)||(headA->data!=headB->data))
return 0;
return CompareLists(headA->next,headB->next); }
sharonalexander1 Python Style :D
def compare_lists(llist1, llist2): if ((llist1 is not None)^(llist2 is not None)): return 0 else: msg = (llist1 is None and llist2 is None)and 1 or (llist1.data==llist2.data and compare_lists(llist1.next,llist2.next) or 0) return msg
zed1g int CompareLists(Node *h1, Node *h2) { while (h1 && h2 && (h1->data == h2->data)) { h1 = h1->next; h2 = h2->next; } return h1 == h2; }
Lo_case Wow! Simple n short. Thank you
arohwedd This is a clean solution and works for this specific situation. My only concern is that it changes values on the list itself. Recursion is the better solution because of this.
cayhan you are supposed to return an integer, not a boolean :)
mkmarathe1 implicit casting from boolean to int is OK. otherway around is bad :-P
MIC0DE if the first data of the 2 lists are different, how does it return 0 since h1 and h2 are both not null?
RodneyShag It luckily works since the values of h1 and h2 will always be different since they store pointer locations. So, technically it returns false, but for a very confusing reason.
vishva_nath Please check this,your solution just got AC but this is not right approch to compare addresses. https://www.hackerrank.com/challenges/compare-two-linked-lists/forum/comments/199610
marianomacchi Nice solution, thanks! I used it to create this recursive version in Python3.
def CompareLists(headA, headB): if (headA and headB) and (headA.data == headB.data): return CompareLists(headA.next, headB.next) else: return 1 if headA == headB else 0
PsychicPhoenix int CompareLists(Node headA, Node headB) { while( headA != null && headB != null &&headA.data == headB.data){ headA = headA.next; headB = headB.next; } return (headA == headB)? 1 : 0; }
Jiban Great idea...nice algorithm
vivek_bj1 Best code, clean solution
meghagulati03 thats pretty cool. great job!
meecheck [deleted]vivek_bj1 After loop ends, there are two possibilities: 1) Either of the list reaches NULL so headA==headB returns false because one of them is pointing to valid memory address. 2) or Both points to NULL so headA==headB returns true, oth pointing to invalid address/memory.
Yes you can consider headA and headB pointing to same memory but NULL means "No memory".
meecheck [deleted]
daniyark [deleted]14p350 return (headA == headB)? 1 : 0; can two nodes compared like this?
android_salnikov [deleted]
EduardoLeggiero Solved with recursion:
int CompareLists(Node headA, Node headB) { if (headA == null && headB == null) { return 1; } if (headA == null || headB == null || headA.data != headB.data) { return 0; } return CompareLists(headA.next, headB.next); }terryjames07 Got the same answer!
int CompareLists(Node a, Node b) { if(a == null && b == null) { return 1; } else if((a == null || b == null) || (a.data != b.data)) { return 0; } return CompareLists(a.next, b.next); }
TheExorcist elegant one
ravan_yvr bro can you explain me using return before calling recursion function
dasdachs It's been a while since my last recursion in Python, but here it goes:
def CompareLists(headA, headB): if not headA or not headB: return int(headA == headB) else: compare_node_data = headA.data == headB.data return int(compare_node_data and CompareLists(headA.next, headB.next))
Edited base case to return an int. Thanks to @limaroche.
liamroche You need a cast on that first return as well: the problem requires a 1 to be returned, not a True.
Eg in Python 2 or Python 3:
def CompareLists(headA, headB): while headA or headB: if not (headA and headB) or headA.data != headB.data: return 0 headA = headA.next headB = headB.next return 1
HHWWHH Another way--With an error handler in python:
def CompareLists(headA, headB): cura=headA curb=headB while cura!=None or curb!=None: try: if cura.data!=curb.data: return 0 else: cura=cura.next curb=curb.next except: return 0 return 1
davidwihl Your code includes an extra check if the lists are different lengths. This way may be a little simpler and not require the exception handler:
def CompareLists(headA, headB): curA = headA curB = headB while curA and curB: if curA.data != curB.data: return 0 curA = curA.next curB = curB.next return 1 if curA == curB else 0
MisterHax int CompareLists(Node headA, Node headB) { if(headA == null || headB == null){ return (headA==null && headB==null) ? 1:0; } return (headA.data == headB.data) ? CompareLists(headA.next, headB.next) : 0; }
sharonalexander1 Python rules:D
def compare_lists(llist1, llist2): if ((llist1 is not None)^(llist2 is not None)): return 0 else: msg = (llist1 is None and llist2 is None)and 1 or (llist1.data==llist2.data and compare_lists(llist1.next,llist2.next) or 0) return msg
eghc4nxohb3p Nice recursive solution with bonus points for the XOR!
auuine function compareLinkedLists( headA, headB) { // 1 Size // 2 Values for (let currentA = headA, currentB = headB; currentA; currentA = currentA.next, currentB = currentB.next) { if (!currentB || !currentA.next && currentB.next) { return 0 } if (currentA.data !== currentB.data) { return 0 } } return 1 }
manoj_nagthane what's problem with this one
int CompareLists(Node *headA, Node* headB) { Node *tempa=headA; Node *tempb=headB; while(tempa!=NULL && tempb!=NULL){ if(tempa->data!=tempb->data){ return 0; } else { return 1;} tempa=tempa->next; tempb=tempb->next; } // This is a "method-only" submission. // You only need to complete this method }
poojaprakash How abt this ?
int CompareLists(Node headA, Node headB) { Node currA = headA; Node currB = headB; int test = 0; while (currA != null && currB != null){ if(currA.data != currB.data){ test = 0; return test; } currA = currA.next; currB = currB.next; } if(currA == null && currB == null) { test = 1; }
return test; }
prajith1 [deleted]
Sort 334 Discussions, By:
Please Login in order to post a comment