LeetCode 88. Merge Sorted Array 解題紀錄
題目
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== n- 0 $\leq$ m, n $\leq$ 200
- 1 $\leq$ m + n $\leq$ 200
- $-10^9 \leq$
nums1[i],nums2[j]$\leq 10^9$
Follow up: Can you come up with an algorithm that runs in $\mathcal{O}(m+n)$ time?
想法
要把所有的值依照 sorted 塞回 nums1 裡，只要比較兩個 array 中最大值就可以知道應該要把誰放進去了。
如果 nums1 全部的值都比 nums2 大，則需要最後一個 while 將所有值放進 nums1。
解法
void merge(vector<int>& nums1, int m, vector<int>& nums2, int n) {
auto i = m-1, j = n-1, pos = m+n-1;
while(i>=0 && j>=0)
{
if(nums1[i] < nums2[j])
{
nums1[pos--] = nums2[j--];
}
else
{
nums1[pos--] = nums1[i--];
}
}
while(j>=0)
{
nums1[pos--] = nums2[j--];
}
}
Time complexity: $\mathcal{O}(m+n)$.
Space complexity: $\mathcal{O}(1)$.