Description:
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.
Examples:
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 in Python:
To solve this problem, we need to merge two sorted arrays, nums1 and nums2, into one sorted array in-place within nums1. The key insight is to start merging from the end of nums1 to avoid overwriting elements that haven’t been processed yet. We’ll use a three-pointer approach: one pointer for nums1, one for nums2, and one for the position in the merged array.
Python
from typing import List
class Solution:
def merge(self, nums1: List[int], m: int, nums2: List[int], n: int) -> None:
"""
Merge nums2 into nums1 in-place. The final sorted array should not be returned
by the function, but instead be stored inside the array nums1.
"""
# Start from the end of nums1 and nums2
p1 = m - 1 # Pointer for the last element in nums1's initial elements
p2 = n - 1 # Pointer for the last element in nums2
p_merge = m + n - 1 # Pointer for the last position in nums1
# Compare elements from nums1 and nums2 and fill nums1 from the end
while p1 >= 0 and p2 >= 0:
if nums1[p1] > nums2[p2]:
nums1[p_merge] = nums1[p1]
p1 -= 1
else:
nums1[p_merge] = nums2[p2]
p2 -= 1
p_merge -= 1
# If there are remaining elements in nums2, place them in nums1
# No need to check for nums1 because they are already in place
while p2 >= 0:
nums1[p_merge] = nums2[p2]
p_merge -= 1
p2 -= 1
# Example usage:
sol = Solution()
nums1 = [1,2,3,0,0,0]
m = 3
nums2 = [2,5,6]
n = 3
sol.merge(nums1, m, nums2, n)
print(nums1) # Output: [1, 2, 2, 3, 5, 6]Explanation:
- Initialization:
p1is initialized tom - 1, which is the index of the last element in the initial part ofnums1.p2is initialized ton - 1, which is the index of the last element innums2.p_mergeis initialized tom + n - 1, which is the index of the last position innums1.
- Merging:
- We iterate through
nums1andnums2from the end, comparing elements and placing the larger one at the current position pointed to byp_merge. - We decrease the appropriate pointers (
p1,p2,p_merge) after each operation.
- We iterate through
- Handling Remaining Elements:
- If
p2still has elements left after the main loop, we copy them intonums1at the appropriate positions. - If
p1has elements left, they are already in their correct positions, so no action is needed.
- If
This approach ensures we don’t overwrite any elements before they are processed and efficiently merges the two arrays in-place.
Solution in Javascript:
JavaScript
/**
* Merges two sorted arrays into one sorted array in-place.
* @param {number[]} nums1 - The first sorted array, which has a length of m + n.
* @param {number} m - The number of initialized elements in nums1.
* @param {number[]} nums2 - The second sorted array.
* @param {number} n - The number of elements in nums2.
* @return {void} Do not return anything, modify nums1 in-place instead.
*/
var merge = function(nums1, m, nums2, n) {
// Initialize three pointers
let p1 = m - 1; // Pointer for the last initialized element in nums1
let p2 = n - 1; // Pointer for the last element in nums2
let pMerge = m + n - 1; // Pointer for the last position in nums1
// Compare elements from the end of nums1 and nums2 and place them in the correct position
while (p1 >= 0 && p2 >= 0) {
if (nums1[p1] > nums2[p2]) {
nums1[pMerge] = nums1[p1];
p1--;
} else {
nums1[pMerge] = nums2[p2];
p2--;
}
pMerge--;
}
// If there are any remaining elements in nums2, copy them over to nums1
while (p2 >= 0) {
nums1[pMerge] = nums2[p2];
pMerge--;
p2--;
}
};
// Example usage:
let nums1 = [1, 2, 3, 0, 0, 0];
let m = 3;
let nums2 = [2, 5, 6];
let n = 3;
merge(nums1, m, nums2, n);
console.log(nums1); // Output: [1, 2, 2, 3, 5, 6]Solution in Java:
Java
class Solution {
public void merge(int[] nums1, int m, int[] nums2, int n) {
// Initialize three pointers
int p1 = m - 1; // Pointer for the last initialized element in nums1
int p2 = n - 1; // Pointer for the last element in nums2
int pMerge = m + n - 1; // Pointer for the last position in nums1
// Compare elements from the end of nums1 and nums2 and place them in the correct position
while (p1 >= 0 && p2 >= 0) {
if (nums1[p1] > nums2[p2]) {
nums1[pMerge] = nums1[p1];
p1--;
} else {
nums1[pMerge] = nums2[p2];
p2--;
}
pMerge--;
}
// If there are any remaining elements in nums2, copy them over to nums1
while (p2 >= 0) {
nums1[pMerge] = nums2[p2];
pMerge--;
p2--;
}
// No need to copy the remaining elements from nums1 since they are already in place
}
public static void main(String[] args) {
Solution sol = new Solution();
int[] nums1 = {1, 2, 3, 0, 0, 0};
int m = 3;
int[] nums2 = {2, 5, 6};
int n = 3;
sol.merge(nums1, m, nums2, n);
System.out.println(java.util.Arrays.toString(nums1)); // Output: [1, 2, 2, 3, 5, 6]
}
}Solution in C#:
C#
using System;
public class Solution {
public void Merge(int[] nums1, int m, int[] nums2, int n) {
// Initialize three pointers
int p1 = m - 1; // Pointer for the last initialized element in nums1
int p2 = n - 1; // Pointer for the last element in nums2
int pMerge = m + n - 1; // Pointer for the last position in nums1
// Compare elements from the end of nums1 and nums2 and place them in the correct position
while (p1 >= 0 && p2 >= 0) {
if (nums1[p1] > nums2[p2]) {
nums1[pMerge] = nums1[p1];
p1--;
} else {
nums1[pMerge] = nums2[p2];
p2--;
}
pMerge--;
}
// If there are any remaining elements in nums2, copy them over to nums1
while (p2 >= 0) {
nums1[pMerge] = nums2[p2];
pMerge--;
p2--;
}
// No need to copy the remaining elements from nums1 since they are already in place
}
}