88. Merge Sorted Array

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88. Merge Sorted Array

Description

Difficulty: Easy

Related Topics: Array, Two Pointers, Sorting

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:

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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:

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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:

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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?

Hints/Notes

Solution

Language: C++

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class Solution {
public:
    void merge(vector<int>& nums1, int m, vector<int>& nums2, int n) {
        int idx1 = m - 1, idx2 = n - 1, idx = m + n - 1;
        while (idx1 >= 0 || idx2 >= 0) {
            if (idx1 < 0) {
                nums1[idx--] = nums2[idx2--];
            } else if (idx2 < 0) {
                nums1[idx--] = nums1[idx1--];
            } else if (nums1[idx1] > nums2[idx2]) {
                nums1[idx--] = nums1[idx1--];
            } else {
                nums1[idx--] = nums2[idx2--];
            }
        }
    }
};