https://leetcode.com/problems/merge-sorted-array/?envType=study-plan-v2&envId=top-interview-150
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.
Solution:
class Solution {
public:
void merge(vector<int>& nums1, int m, vector<int>& nums2, int n) {
int i = m - 1;
int j = n - 1;
int k = m + n - 1;
while(j >= 0) {
cout << j << endl;
if(i >= 0 && nums1[i] > nums2[j]) {
nums1[k] = nums1[i];
k--;
i--;
} else {
nums1[k] = nums2[j];
k--;
j--;
}
}
}
};