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
Follow up: Can you come up with an algorithm that runs in O(m + n) time?
class Solution {
public void merge(int[] nums1, int m, int[] nums2, int n) {
int i = m-1;
int j = n-1;
int k = m+n-1;
while(i>=0||j>=0){
if(i<0){
nums1[k--] = nums2[j--];
}else if(j<0){
nums1[k--] = nums1[i--];
}else if(nums1[i]<nums2[j]){
nums1[k--] = nums2[j--];
}else{
nums1[k--] = nums1[i--];
}
}
}
}class Solution {
public void merge(int[] nums1, int m, int[] nums2, int n) {
int tail1 = m - 1, tail2 = n - 1, icur = m + n - 1;
while (tail1 >= 0 && tail2 >= 0) {
nums1[icur--] = (nums1[tail1] > nums2[tail2]) ? nums1[tail1--] : nums2[tail2--];
}
while (tail2 >= 0) { //only need to combine with remaining nums2
nums1[icur--] = nums2[tail2--];
}
}
}从后往前