(Leet Code c++)Merge Sorted Array

88. Merge Sorted Array

 

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

주어진 num1과 num2는 오름차순으로 정렬되어있다.

num1의 원소 중 m개, num2의 원소 중 n개를 v 벡터에 저장하여 오름차순으로 정렬한다.

그후, v를 다시 num1에 대입하면 끝!

class Solution {
public:
    void merge(vector<int>& nums1, int m, vector<int>& nums2, int n) {
        vector<int> v;
        
        for(int i = 0; i < nums1.size(); i++){
            if(m > 0){
                m--;
                v.push_back(nums1[i]);
            }
            else{
                break;
            }
        }
        
        for(int i = 0; i < nums2.size(); i++){
            if(n > 0){
                n--;
                v.push_back(nums2[i]);
            }
            else{
                break;
            }
        }
        
        sort(v.begin(), v.end());
        nums1 = v;
    }
};