Today I have done Reverse Nodes in k-Group and leetcode's January LeetCoding Challenge with cpp.

Reverse Nodes in k-Group

Description

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

k is a positive integer and is less than or equal to the length of the linked list. If the number of nodes is not a multiple of k then left-out nodes, in the end, should remain as it is.

Follow up:

• Could you solve the problem in O(1) extra memory space?
• You may not alter the values in the list's nodes, only nodes itself may be changed.

Example 1:

Input: head = [1,2,3,4,5], k = 2
Output: [2,1,4,3,5]

Example 2:

Input: head = [1,2,3,4,5], k = 3
Output: [3,2,1,4,5]

Example 3:

Input: head = [1,2,3,4,5], k = 1
Output: [1,2,3,4,5]

Example 4:

Input: head = [1], k = 1
Output: [1]

Constraints:

• The number of nodes in the list is in the range sz.
• 1 <= sz <= 5000
• 0 <= Node.val <= 1000
• 1 <= k <= sz

Solution

why this problem is a hard?

maybe I should continue to improve solution?

class Solution {
    int length(ListNode *head) {
        int len = 0;
        while(head) {
            len += 1;
            head = head->next;
        }
        return len;
    }
public:
    ListNode* reverseKGroup(ListNode* head, int k) {
        ListNode *newHead = new ListNode();
        ListNode *cur = head, *tail = newHead;
        int len = length(head);
        if(k == 1 || len < k) return head;
        while(len >= k) {
            int cnt = k;
            ListNode *partHead = nullptr;
            ListNode *next;
            while(cnt--) {
                next = cur->next;
                tail->next = cur;
                cur->next = partHead;
                partHead = cur;
                cur = next;
            }
            while(tail->next) tail = tail->next;
            len -= k;
        }
        tail->next = cur;
        return newHead->next;
    }
};

January LeetCoding Challenge 11

Description

Merge Sorted Array

Given two sorted integer arrays nums1 and nums2, merge nums2 into nums1 as one sorted array.

The number of elements initialized in nums1 and nums2 are m and n respectively. You may assume that nums1 has enough space (size that is equal to m + n) to hold additional elements from nums2.

Example 1:

Input: nums1 = [1,2,3,0,0,0], m = 3, nums2 = [2,5,6], n = 3
Output: [1,2,2,3,5,6]

Example 2:

Input: nums1 = [1], m = 1, nums2 = [], n = 0
Output: [1]

Constraints:

• 0 <= n, m <= 200
• 1 <= n + m <= 200
• nums1.length == m + n
• nums2.length == n
• -109 <= nums1[i], nums2[i] <= 109

Solution

class Solution {
public:
    void merge(vector<int>& nums1, int m, vector<int>& nums2, int n) {
        while(n && m) {
            if(nums1[m-1] > nums2[n-1]) {
                nums1[n+m-1] = nums1[m-1];
                --m;
            } else {
                nums1[n+m-1] = nums2[n-1];
                --n;
            }
        }
        while(n) {
            nums1[n-1] = nums2[n-1];
            --n;
        }
    }
};