[LeetCode 88] Merge Sorted Array

문제 바로가기

Description

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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

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• nums1.length == m + n
• nums2.length == n
• 0 <= m, n <= 200
• 1 <= m + n <= 200
• -10^9 <= nums1[i], nums2[j] <= 10^9

Hint

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Hint 1

  You can easily solve this problem if you simply think about two elements at a time rather than two arrays. We know that each of the individual arrays is sorted. What we don't know is how they will intertwine. Can we take a local decision and arrive at an optimal solution?
	

Hint 2

  If you simply consider one element each at a time from the two arrays and make a decision and proceed accordingly, you will arrive at the optimal solution.
	

Code

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내 제출

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class Solution {
    public void merge(int[] nums1, int m, int[] nums2, int n) {
        List list = new ArrayList();

        for (int i = 0; i < m; i++) {
            list.add(nums1[i]);
        }

        for (int i = 0; i < n; i++) {
            list.add(nums2[i]);
        }

        Collections.sort(list);

        for (int i = 0; i < list.size(); i++) {
            nums1[i] = (int)list.get(i);
        }
    }
}

Runtime	Memory
1 ms	42.2 MB

다른 풀이

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class Solution {
    public void merge(int[] nums1, int m, int[] nums2, int n) {
        int i =m-1;
        int j =n-1;
        int k=m+n-1;
        while(i>=0 && j>=0){
            if(nums1[i]>nums2[j]){
                nums1[k]=nums1[i];
                i--;
            }else{
                nums1[k]=nums2[j];
                j--;
            }
            k--;
        }
        while(j>=0){
            nums1[k]=nums2[j];
            j--;
            k--;
        }
        
    }
}

Reference

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