88. Merge Sorted Array

Question

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.

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.

Constraints:

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

Algorithm

1. Read question carefully to understand what's the requirements;
2. Handle edge case; (thanks to the examples)
3. Start from the end then you can do the comparison and set in one pass, because the last n digits of nums1 has been pre occupied for you to set new values

Code

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