Description
https://leetcode.com/problems/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 + nnums2.length == n0 <= m, n <= 2001 <= m + n <= 200-109 <= nums1[i], nums2[j] <= 109
Follow up: Can you come up with an algorithm that runs in O(m + n) time?
Explanation
The size of nums1 is m + n. nums1 is enough to store all numbers in nums1 and nums2. We only need to first shift all numbers in nums1 by n positions. Then we compare and put numbers in nums1 and nums2 into nums1 in ascending order.
Python Solution
class Solution:
def merge(self, nums1: List[int], m: int, nums2: List[int], n: int) -> None:
"""
Do not return anything, modify nums1 in-place instead.
"""
for i in range(m - 1, -1, -1):
nums1[i + n] = nums1[i]
i = n
j = 0
k = 0
while i < m + n and j < n:
if nums1[i] < nums2[j]:
nums1[k] = nums1[i]
i += 1
else:
nums1[k] = nums2[j]
j += 1
k += 1
while j < n:
nums1[k] = nums2[j]
j += 1
k += 1- Time complexity: O(m + n).
- Space complexity: O(m + n).
class Solution {
public void merge(int[] nums1, int m, int[] nums2, int n) {
for (int i = m; i < (m+n); i++) {
nums1[i] = nums2[i – m];
}
Arrays.sort(nums1);
}
}
/** Java solution */
public void merge(int[] nums1, int m, int[] nums2, int n) {
int insertP = m + n – 1;
int nums1P = m – 1;
int nums2P = n – 1;
while (nums1P >= 0 && nums2P >= 0){
if (nums1[nums1P] > nums2[nums2P]) {
nums1[insertP–] = nums1[nums1P–];
}
else {
nums1[insertP–] = nums2[nums2P–];
}
}
while (nums2P >= 0) {
nums1[insertP–] = nums2[nums2P–];
}
}