難度: Easy
類型: Array,Two Pointers,Sorting
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
Consideration:
If starting from left to right, the merged items may overwrite nums1 elements. Therefore, we start to compare the larger item and fill from right to left. It may save time and memory to re-arrange elements to avoid overwriting.
Code:
class Solution {
public:
void merge(vector<int>& nums1, int m, vector<int>& nums2, int n) {
int i, j=m-1, k=n-1;
for (i=m+n-1;i>=0;i--)
{
//cout << "i = " << i << "\n";
//cout << "m = " << m << "\n";
//cout << "n = " << n << "\n";
if ( j >= 0 && k >= 0 )
{
if ( nums1.at(j) >= nums2.at(k) )
{
nums1.at(i) = nums1.at(j);
j--;
}
else
{
nums1.at(i) = nums2.at(k);
k--;
}
}
else if (j<0)
{
nums1.at(i) = nums2.at(k);
k--;
}
else if (k<0)
{
nums1.at(i) = nums1.at(j);
j--;
}
else
{
cout << "Error! Check the logic!!";
}
}
}
};
Result:
