我在哪里 go 错误地合并两个排序的 arrays。 它们需要在 O(1) 空间复杂度中合并

Question

I am sorry. I am new to this forum and also to programming. I have seen a question Merge without extra space and I tried solving it. I haven't seen the editorial. I have tried writing the solution for the code, but it didn't seem to work online. Although I tried a few test cases of my own which seem to work on my native ide. I want to know if my code is even right and I want to know how can I make it right.

Here's my code:

void merge(int arr1[], int arr2[], int n, int m) {  // n is the size of arr1, m is size of arr2
    int i=0;
    while(i<n){
        if (arr1[i]>arr2[0])
        {
            swap(arr1[i], arr2[0]);
        }
        for (int j=1; j<m; j++){
            if (arr2[j-1]>arr2[j]){
                swap(arr2[j-1], arr2[j]);
            }
        }
        i++;
    }
}

I want the space complexity to be O(1) and the expected time complexity to be O((n+m) log(n+m)).

Answer 1

With that time complexity of O((n+m) log(n+m) you might do regular sort.

Creating correct iterator can be done but not trivial.

Other solution is to first partition elements to put the n lowest elements in arr1 , then sort each part:

void merge(int arr1[], int arr2[], int n, int m) {
    int i1 = 0;
    int i2 = 0;

    for (int i = 0; i != n; ++i) { // O(n)
        if (arr[i1] < arr2[i2]) { ++i1; }
        else { ++i2; }
    }
    // i1 + i2 == n
    for (int i = 0; i != i2; ++i) { // O(n)
        std::swap(arr2[i], arr1[i1 + i]);
    }

    std::sort(arr1, arr1 + n); // O(n log(n))
    std::sort(arr2, arr2 + m); // O(m log(m))
}

Demo .

• O(n log(n)) < O(n log(n + m))
• O(m log(m)) < O(m log(n + m))
• O(n log(n)) + O(m log(m)) < O((n + m) log(n + m))

Note: I am even confident there exists a solution in O(n+m) without extra place.