Reverse Nodes in k-Group

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Given a linked list, reverse the nodes of a linked list k at a time and return its modified list.

If the number of nodes is not a multiple of k then left-out nodes in the end should remain as it is.

You may not alter the values in the nodes, only nodes itself may be changed.

Only constant memory is allowed.

For example,
Given this linked list: 1->2->3->4->5

For k = 2, you should return: 2->1->4->3->5

For k = 3, you should return: 3->2->1->4->5

class Solution {

    void revRec(ListNode *pre, int k){
        ListNode *end = pre;
        for(int i = 0; i < k; i++){
            if(end->next == NULL)    return;
            end = end->next;
//when k==1, nextPre will be the end and do not go into the while
        ListNode *nextPre = end;
        while(pre->next != end){
            ListNode *start = pre->next;            
            if(nextPre == end) nextPre = start;
            pre->next = pre->next->next;
            start->next = end->next;
            end->next = start;
        revRec(nextPre, k);

    ListNode *reverseKGroup(ListNode *head, int k) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        //if(head == 0)   return NULL;
        ListNode *dummy = new ListNode(0);
        dummy->next = head;
        ListNode *pre = dummy;
        revRec(pre, k);
        ListNode *ret = dummy->next;
        delete dummy;
        return ret;
class Solution {

    ListNode* reverseOneGroup(ListNode* pre, ListNode* last) {
        ListNode* newLast = pre->next;
        while (pre->next != last) {
            ListNode* p1 = pre->next;
            pre->next = p1->next;
            p1->next = last->next;
            last->next = p1;
        return newLast;

    ListNode *reverseKGroup(ListNode *head, int k) {
        if (head == NULL)   return NULL;
        ListNode* dummy = new ListNode(0);
        dummy->next = head;
        ListNode* pre = dummy;
        while (true) {
            ListNode* last = pre;
            for (int i = 0; i < k; i++) {
                if (last->next) {
                    last = last->next;
                } else {
                    return dummy->next;
            pre = reverseOneGroup(pre, last);

Written by linzhongzl

April 24, 2013 at 11:16 am

Posted in Leetcode

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