88

88. Merge Sorted Array (Easy)

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

 

Follow up: Can you come up with an algorithm that runs in O(m + n) time?

Related Topics:
Array, Two Pointers, Sorting

Similar Questions:

• Merge Two Sorted Lists (Easy)
• Squares of a Sorted Array (Easy)
• Interval List Intersections (Medium)

Solution 1.

// OJ: https://leetcode.com/problems/merge-sorted-array/
// Author: github.com/lzl124631x
// Time: O(M + N)
// Space: O(1)
class Solution {
public:
    void merge(vector<int>& A, int m, vector<int>& B, int n) {
        for (int i = m - 1, j = n - 1, k = m + n - 1; k >= 0; --k) {
            if (j < 0 || (i >= 0 && A[i] > B[j])) A[k] = A[i--];
            else A[k] = B[j--];
        }
    }
};