88. Merge Sorted Array

1. Description

You are given two integer arrays nums1 and nums2, sorted in non-decreasing order, and two integers m and n, representing the number of elements in nums1 and nums2 respectively.
Merge nums1 and nums2 into a single array sorted in non-decreasing order.
The final sorted array should not be returned by the function, but instead be stored inside the array nums1. To accommodate this, nums1 has a length of m + n, where the first m elements denote the elements that should be merged, and the last n elements are set to 0 and should be ignored. nums2 has a length of n.

2. Example

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]
Explanation: The arrays we are merging are [1,2,3] and [2,5,6].
The result of the merge is [1,2,2,3,5,6] with the underlined elements coming from nums1.

Example 2

Input: nums1 = [1], m = 1, nums2 = [], n = 0
Output: [1]
Explanation: The arrays we are merging are [1] and [].
The result of the merge is [1].

Example 3

Input: nums1 = [0], m = 0, nums2 = [1], n = 1
Output: [1]
Explanation: The arrays we are merging are [] and [1].
The result of the merge is [1].
Note that because m = 0, there are no elements in nums1. The 0 is only there to ensure the merge result can fit in nums1.

3. Constraints

• nums1.length == m + n
• nums2.length == n
• 0 <= m, n <= 200
• 1 <= m + n <= 200
• -10$^{9}$ <= nums1[i], nums2[j] <= 10$^{9}$

4. Solutions

Two Pointers

Time complexity: O(m + n)
Space complexity: O(1)

class Solution {
public:
    void merge(vector<int> &nums1, int m, vector<int> &nums2, int n) {
        for (int i = m + n - 1; i >= 0 && n > 0; --i) {
            if (m == 0 || nums1[m - 1] <= nums2[n - 1]) {
                nums1[i] = nums2[n - 1];
                --n;
            } else {
                nums1[i] = nums1[m - 1];
                --m;
            }
        }
    }
};