LeetCode 솔루션

[Easy - wk3 - Q3] 88. Merge Sorted Array

Chloe 쪽지보내기 메일보내기 자기소개 아이디로 검색 전체게시물 작성
작성일 2022.05.16 17:02

7,006 조회
3 댓글
0 추천
0 비추천
목록 게시글 쓰기

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?

#leetcode, #문제풀이, #easy, #facebook, #microsoft, #apple, #linkedin, #amazon, #bloomberg, #shopee, #indeed, #uber, #oracle, #splunk, #array, #two pointers, #sorting

링크

https://leetcode.com/problems/merge-sorted-array/ 4212 회 연결

익명
작성일 2022.05.18 19:45

python

Runtime: 30 ms, faster than 98.58% of Python3 online submissions for Merge Sorted Array.
Memory Usage: 13.9 MB, less than 38.93% of Python3 online submissions for Merge Sorted Array

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.
        """
        p1 = m - 1
        p2 = n - 1
        
        for i in range(m+n-1,-1,-1):
            if p2 < 0:
                break
            if p1 >= 0 and nums1[p1] > nums2[p2]:
                nums1[i] = nums1[p1]
                p1 -= 1
            else:
                nums1[i] = nums2[p2]
                p2 -= 1

익명
작성일 2022.05.24 11:28

C++
Runtime: 0 ms, faster than 100.00% of C++ online submissions for Merge Sorted Array.
Memory Usage: 9 MB, less than 90.31% of C++ online submissions for Merge Sorted Array.

class Solution {
public:
    void merge(vector<int>& nums1, int n1, vector<int>& nums2, int n2) {
        int size = n1-- + n2-- - 1;
        
        while (n1 >= 0 || n2 >= 0) {
            if (n1 < 0) nums1[size--] = nums2[n2--];
            else if (n2 < 0) nums1[size--] = nums1[n1--];
            else nums1[size--] = nums1[n1] > nums2[n2] ? nums1[n1--] : nums2[n2--];
        }
    }
};

익명
작성일 2022.05.24 23:26

JS
Runtime: 67 ms, faster than 77.95% of JavaScript online submissions for Merge Sorted Array.
Memory Usage: 41.9 MB, less than 86.92% of JavaScript online submissions for Merge Sorted Array.

var merge = function(nums1, m, nums2, n) {
    // nums1 = [1,2,3,0,0,0], m = 3, nums2 = [2,5,6], n = 3
    // output = [1,2,2,3,5,6]
    let first = m-1;
    let second = n-1;
    let i = m+n-1;
    
    
    //퍼스트가 세컨보다 작다는 커디션을 먼저 적을때 에러 뜸
    // while (second >= 0) {
    //     if (nums1[first] < nums2[second]) {
    //         nums1[i] = nums2[second];
    //         second--;
    //         i--;
    //     } else {
    //         nums1[i] = nums1[first];
    //         first--;
    //         i--;
    //     }
    //}
    
      while (second >= 0) {
          let fVal = nums1[first];
          let sVal = nums2[second];
        if (fVal > sVal) {
            nums1[i] = fVal;
            first--;
            i--;
        } else {
            nums1[i] = sVal;
            second--;
            i--;
        }
    }

JS
Runtime: 67 ms, faster than 77.95% of JavaScript online submissions for Merge Sorted Array.
Memory Usage: 41.9 MB, less than 86.92% of JavaScript online submissions for Merge Sorted Array.
[code=js]var merge = function(nums1, m, nums2, n) {
    // nums1 = [1,2,3,0,0,0], m = 3, nums2 = [2,5,6], n = 3
    // output = [1,2,2,3,5,6]
    let first = m-1;
    let second = n-1;
    let i = m+n-1;
    
    
    //퍼스트가 세컨보다 작다는 커디션을 먼저 적을때 에러 뜸
    // while (second >= 0) {
    //     if (nums1[first] < nums2[second]) {
    //         nums1[i] = nums2[second];
    //         second--;
    //         i--;
    //     } else {
    //         nums1[i] = nums1[first];
    //         first--;
    //         i--;
    //     }
    //}
    
      while (second >= 0) {
          let fVal = nums1[first];
          let sVal = nums2[second];
        if (fVal > sVal) {
            nums1[i] = fVal;
            first--;
            i--;
        } else {
            nums1[i] = sVal;
            second--;
            i--;
        }
    }[/code]