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:
def merge(self, nums1: List[int], m: int, nums2: List[int], n: int) -> None:
"""
Do not return anything, modify nums1 in-place instead.
"""
i, j, k = m - 1, n - 1, m + n - 1
while j >= 0:
if i >= 0 and nums1[i] > nums2[j]:
nums1[k] = nums1[i]
i -= 1
else:
nums1[k] = nums2[j]
j -= 1
k -= 1
class Solution:
def merge(self, nums1: List[int], m: int, nums2: List[int], n: int) -> None:
"""
Do not return anything, modify nums1 in-place instead.
"""
i, j, k = m - 1, n - 1, m + n - 1
while j >= 0:
if i < 0 or nums1[i] < nums2[j]:
nums1[k] = nums2[j]
j -= 1
else:
nums1[k] = nums1[i]
i -= 1
k -= 1
class Solution {
public void merge(int[] nums1, int m, int[] nums2, int n) {
int i = m - 1, j = n - 1, k = m + n - 1;
while (j >= 0) {
if (i >= 0 && nums1[i] > nums2[j]) {
nums1[k--] = nums1[i--];
} else {
nums1[k--] = nums2[j--];
}
}
}
}
func merge(nums1 []int, m int, nums2 []int, n int) {
i, j, k := m - 1, n - 1, m + n - 1
for j >= 0 {
if i >= 0 && nums1[i] > nums2[j] {
nums1[k] = nums1[i]
i--
} else {
nums1[k] = nums2[j]
j--
}
k--
}
}
/**
* @param {number[]} nums1
* @param {number} m
* @param {number[]} nums2
* @param {number} n
* @return {void} Do not return anything, modify nums1 in-place instead.
*/
var merge = function (nums1, m, nums2, n) {
let i = m - 1,
j = n - 1,
k = m + n - 1;
while (j >= 0) {
if (i >= 0 && nums1[i] > nums2[j]) {
nums1[k--] = nums1[i--];
} else {
nums1[k--] = nums2[j--];
}
}
};
impl Solution {
pub fn merge(nums1: &mut Vec<i32>, m: i32, nums2: &mut Vec<i32>, n: i32) {
let (mut m, mut n) = (m as usize, n as usize);
for i in (0..m + n).rev() {
nums1[i] = match (m == 0, n == 0) {
(true, false) => {
n -= 1;
nums2[n]
}
(false, true) => {
m -= 1;
nums1[m]
}
(_, _) => {
if nums1[m - 1] > nums2[n - 1] {
m -= 1;
nums1[m]
} else {
n -= 1;
nums2[n]
}
}
}
}
}
}