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.
My solution
class Solution {
public:
void merge(vector<int>& nums1, int m, vector<int>& nums2, int n) {
vector<int> temp(m+n);
int i=0,j=0;
if (n==0) return;
for(int l = 0;l<m+n;l++){
if (i==m){
temp[l] = nums2[j];
j++;
continue;
}
if(j==n){
temp[l] = nums1[i];
i++;
continue;
}
if (nums1[i]>nums2[j]){
temp[l] = nums2[j];
j++;
}
else{
temp[l] = nums1[i];
i++;
}
}
nums1 = temp;
}
};
Optimal Solution
O(1) in memory
class Solution {
public:
void merge(vector<int>& nums1, int m, vector<int>& nums2, int n) {
int i = m - 1; // Last element in nums1's valid part
int j = n - 1; // Last element in nums2
int k = m + n - 1; // Last position in nums1
while (i >= 0 && j >= 0) {
if (nums1[i] > nums2[j]) {
nums1[k--] = nums1[i--];
} else {
nums1[k--] = nums2[j--];
}
}
// Copy remaining elements from nums2 (if any)
while (j >= 0) {
nums1[k--] = nums2[j--];
}
// No need to copy nums1's remaining elements - they're already in place
}
};