@weby-vansh
we can also use this recursive approach :
class Solution {
public:
ListNode* swapPairs(ListNode* head) {
int k = 2;
ListNode* temp = head;
int count = 0;
// Step 1: Check if we have at least k nodes
while (count < k && temp != nullptr) {
temp = temp->next;
count++;
}
// If we have fewer than k nodes, return head (no reversal)
if (count < k) return head;
// Step 2: Reverse k nodes
ListNode* prev = swapPairs(temp); // Recurse for the rest of the list
temp = head;
int cnt = 0;
while (cnt < k) {
ListNode* next = temp->next;
temp->next = prev;
prev = temp;
temp = next;
cnt++;
}
return prev;
}
};
@GurudasDev
i have done this question on my own i have finally succeed
@rish_tr
I accomplished it! I just had to watch the last part once. This part:
while (count < k) {
ListNode* next = temp->next;
temp->next = prevNode;
prevNode = temp;
temp = next;
count++;
}
@ShushantKumar-y2u
Your way of explaining is awesome Didi, You explain difficult questions in a very simple way.
@SHQUIZCORNER
React js course when coming plz replay......
@OMEGAGaming-cv1ty
please continue the DSA series
@tanishqvishwakarma9518
Done 👍 Thank you Didi 😊
@SonaliParida-q5c
Awsome explaination
@samyabiswas8366
You're an awesome educator di!!
@MOHD_ASHRAF.707
Great lecture 🥰📌🥰📌🥰🔥🔥🔥📌📌🥰🥰🥰🥰📌
@HayAbhi1
Continuity go slowly, But go High ❤
@PrashChandra71
Done Today's Lecture❤
@vivekmalviya3124
Thanku you very much for this DSA series
@sumanexclusive5653
thank you didi ...😍😍😍
@DhananjayIngole-jy3rh
plz continue the series plz
@Shaswat_Shrivas
Lecture Completed!
@sharma-q2q
Hello Didi , i'm from Bangladesh . Since you began creating videos for the DSA course using C + +, I have been viewing your videos on a regular basis. I also gained a lot of fresh knowledge. If I wish to enrol in your course, what can I do?
@hemadevi7106
This question can be solved with same algorithm discuss in previous video(leetcode 25) with k=2.. Relpy if and edge case stucks😊😊😊
@learn_with_rks
Please upload the next video
@shradhaKD