Merge Sorted Array

Problem

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
$-10^9$ <= nums1[i], nums2[j] <= $10^9$

Follow up:

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

Solution

• vector STL 사용하기
• i, j, k 포인터를 사용하여 큰 숫자부터 nums1에 추가한다.
• 앞에서 부터 할당하면 nums1의 숫자가 삭제된다.
• j 인덱스를 먼저 체크하는 이유는, nums1에 숫자가 하나도 없을 경우에는 i=0-1이 되기때문이다.

Code

#include <iostream>
#include <vector>
using namespace std;

class Solution {
public:
    void merge(vector<int>& nums1, int m, vector<int>& nums2, int n) {

		int i = m - 1;
		int j = n - 1;
		int k = m + n - 1;

		while (j >= 0) {
			if (i >= 0 && nums1[i] > nums2[j]) {
				nums1[k--] = nums1[i--];
			}
			else {
				nums1[k--] = nums2[j--];
			}
		}
    }
	
	void merge2(vector<int>& nums1, int m, vector<int>& nums2, int n) {

		for (int i = m, j = 0; j < n; j++) {
			nums1[i] = nums2[j];
			i++;
		}
		sort(nums1.begin(), nums1.end()); // sort(first, last)
	}
};