Reference: LeetCode
Difficulty: Easy

My Post: Java Solution with Clean Code (Two Pointers)

Problem

Given two sorted integer arrays nums1 and nums2, merge nums2 into nums1 as one sorted array.

Note:

  • The number of elements initialized in nums1 and nums2 are m and n respectively.
  • You may assume that nums1 has enough space (size that is greater or equal to m + n) to hold additional elements from nums2.

Example: 

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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]

Follow up: Can you solve in $O(N)$ without extra space?

Analysis

Merge and Sort

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

Time: $O((m + n)\log{(m + n)})$
Space: $O(1)$

Two Pointers

Like merging two sorted linked list, we can loop over from the beginning of the array. It turns out that we need an extra array of size $m$.

We can start from the end of two arrays and append elements from the right side.

Note: Prefer while-loop to for-loop.

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

Time: $O(m + n)$
Space: $O(1)$