Merge Sorted Array[E,Array,Two Pointers,Sorting]

https://leetcode.com/problems/merge-sorted-array/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?

 

문제부터 넘 길어서 맘에 안 든다.

 

그래도 Merge Sort부분은 항상 중요한 녀석이라 잘 살펴보아야 하지

 

1. Python

PASS는 했는데, 누가봐도 꼼수인 답안.. 다시 생각해보자

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.
        """
        for i in range(m,len(nums1)):
            nums1[i]=nums2[i-m]
        nums1.sort()

 

 

이렇게 했는데, nums1 덮어쓰기가 안 된다..

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.
        """
        num = []
        n1=0
        n2=0
        while n1<m and n2<n :
            if nums1[n1] < nums2[n2]:
                num.append(nums1[n1])
                n1+=1
            else :
                num.append(nums2[n2])
                n2+=1                
        if n1==m:
            num += nums2[n2:]
        else:
            num += nums1[n1:m]
        nums1 = num

 

요런식으로 하면 merge sort가 된다.

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
        p = m+n-1

        while p1>=0 and p2>=0:
            if nums1[p1]<nums2[p2] :
                nums1[p] = nums2[p2]
                p2+= -1
                p+= -1
            else:
                nums1[p] = nums1[p1]
                p1+= -1
                p+= -1
        if p2<0:
            nums1[:p+1]=nums1[:p1+1]
        else:
            nums1[:p+1]=nums2[:p2+1]

 

 

2. C

void merge(int* nums1, int nums1Size, int m, int* nums2, int nums2Size, int n) {
    // if(m==0){
    //     nums1 = nums2;
    //     return;
    // }

    int p1 = m-1;
    int p2 = n-1;
    int p = m+n-1;

    while(p1>=0 && p2 >=0){
        if(nums1[p1] > nums2[p2]){
            nums1[p--] = nums1[p1--];
        }else{
            nums1[p--] = nums2[p2--];
        }
    }
    if(p2<0){    
        for(int i=0;i<p+1;i++){
            nums1[i] = nums1[i];
        }
    }
    else{
        for(int j=0;j<p+1;j++){
            nums1[j] = nums2[j];
        }
    }
}

 

마지막 if/else 구문이 사실은 m or n 이 0인 case도 다 커버해주고 있었던 것!

 

3. C++

class Solution {
public:
    void merge(vector<int>& nums1, int m, vector<int>& nums2, int n) {
        for(int i=m;i<m+n;i++){
            nums1[i]=nums2[i-m];
        }
        sort(nums1.begin(),nums1.end());
    }
};

C++은 분노의 sort 사용

rbegin / rend를 쓰면 reverse sort가 가능하다.