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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.

 

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?

Solutions

Python3

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

Java

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--];
            }
        }
    }
}

Go

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--
    }
}

JavaScript

/**
 * @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--];
        }
    }
};

Rust

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]
                    }
                }
            }
        }
    }
}

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