[LeetCode] #88. Merge Sorted Array (repeat : 0)

두 개의 배열을 합치고 , 오름차순 정렬로 반환하는 문제이다.  sort 메소드를 사용해서 간단히 해결했다. 

 

# Description

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.

 

 

#Solution 

class Solution{
   public void merge(int[] nums1, int m, int[] nums2, int n) {
   	
    	for(int i=0;i<n;i++){
        	nums1[m] = nums2[i];
            m++;
        }
   			Arrays.sort(num1);
   }
}

 

This algorithm uses only two constant variables num1 and num2 , therefore in terms of space complexity, it is O(1).

In terms of time complexity, it is O(n) because the main operation involves in iterating num2 array. It copies each element to the num1 array. so the 'n' represents number of elements in the nums2 array