In the Leetcode Merge Sorted Array problem solution, 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
Solution in C Programming
void merge(int* nums1, int nums1Size, int m, int* nums2, int nums2Size, int n) {
int end = m + n - 1;
while (m > 0 || n > 0) {
int x = m ? nums1[m - 1] : INT_MIN;
int y = n ? nums2[n - 1] : INT_MIN;
if (x > y) {
nums1[end--] = x;
m--;
} else {
nums1[end--] = y;
n--;
}
}
}Solution in C++ Programming
class Solution
{
public:
void merge(vector<int>& nums1, int m, vector<int>& nums2, int n)
{
vector<int>v(n+m);
int i=0,j=0,k=0;
while(i<m && j<n)
{
v[k++]=nums1[i]<nums2[j]?nums1[i++]:nums2[j++];
}
while(i<m)
{
v[k++]=nums1[i++];
}
while(j<n)
{
v[k++]=nums2[j++];
}
i=0,k=0;
while(i<m+n)
{
nums1[i++]=v[k++];
}
}
};Solution in Java Programming
class Solution {
public void merge(int[] nums1, int m, int[] nums2, int n) {
for (int k = m-- + n-- - 1; k >= 0; k--) {
nums1[k] = n < 0 || (m >= 0 && nums1[m] > nums2[n]) ? nums1[m--] : nums2[n--];
}
}
}Solution in Python Programming
class Solution(object):
def merge(self, nums1, m, nums2, n):
a = m - 1
b = n - 1
c = m + n -1
while(b >= 0):
if a >= 0 and nums1[a] >= nums2[b]:
nums1[c] = nums1[a]
a -= 1
else:
nums1[c] = nums2[b]
b -= 1
c -= 1Solution in C# Programming
public class Solution {
public void Merge(int[] nums1, int m, int[] nums2, int n) {
if (nums1.Length < m+n) return;
if (n == 0) return;
//two pointers approach
// start from the end, because both of these arrays are sorted already
int first = m - 1;
int second = n - 1;
for(int k = m+n-1; k>=0; k--) {
if (second < 0) break;
if (first < 0) {
nums1[k] = nums2[second--];
continue;
}
if (nums1[first] > nums2[second]) {
nums1[k] = nums1[first];
first--;
} else {
nums1[k] = nums2[second];
second--;
}
}
}
}Also read,