运行时错误：在 C++ 中合并两个排序的 arrays

Question

I wrote this piece of code for merging two sorted arrays:-

   class Solution {
   public:
     void merge(vector<int>& nums1, int m, vector<int>& nums2, int n) {
         int p1=m-1, p2=n-1, i=m+n-1;      
         while(p2>=0){
     
        if(p1>=0 && nums1[p1]<nums2[p2]){
            
            nums1[i]=nums2[p2];
            p2--;
            
        }//if
        else{
             nums1[i]=nums1[p1];
            p1--;
        }
        i--;
    }
    
}//fn
};

This is throwing up a runtime error. Can someone help me resolve this?

Answer 1

I put OPs sample into an MCVE on coliru and was able to reproduce the issue of OP. In my case, the execution simply was aborted due to an out-of-bounds access.

I highly suspect that OP passed in the nums1, nums1.size(), nums2, nums2.size() into Solution::merge() . As the merge is destructive (writes result into one of the input containers), this has to be considered.

IMHO, OP seems to be aware of this basically as the merge is done from back to front.

However, as nums1 is used for output it has to be provided with sufficient size before merging.

Eg nums1.resize(m + n); at the begin of merge will do the job.

My MCVE with fix:

#include <iostream>
#include <vector>

class Solution {
    
  public:
  
    void merge(std::vector<int>& nums1, int m, std::vector<int>& nums2, int n)
    {
      nums1.resize(m + n); // <= ENSURE SUFFICIENT STORAGE IN nums1
      int p1 = m - 1, p2 = n - 1, i = m + n - 1;      
      while (p2 >= 0) {
        if (p1 >= 0 && nums1[p1] < nums2[p2]) {
          nums1[i] = nums2[p2];
          p2--;
        } else {
          nums1[i] = nums1[p1];
          p1--;
        }
        i--;
      }
    }
};

int main()
{
  // sample data
  std::vector<int> nums1{ 1, 3, 5, 7, 9 };
  std::vector<int> nums2{ 2, 4, 6, 8 };
  // run merge
  Solution().merge(nums1, (int)nums1.size(), nums2, (int)nums2.size());
  // output result
  std::cout << "nums1: {";
  const char* sep = " ";
  for (const int num : nums1) {
    std::cout << sep << num;
    sep = ", ";
  }
  std::cout << " }" << std::endl;
}

Output:

nums1: { 1, 2, 3, 4, 5, 6, 7, 8, 9 }

Live Demo on coliru

---

@Armin Montigny recommended the use of std::merge() instead of a hand-knitted implementation.

First I had some doubts whether it's able to manage destructive merging as well but after thinking twice I realized that it will do it perfectly if used right. I modified my MCVE to see how this would look like.

My alternative MCVE using std::merge() :

#include <algorithm>
#include <functional>
#include <iostream>
#include <vector>

class Solution {
    
  public:
  
    void merge(std::vector<int>& nums1, int m, std::vector<int>& nums2, int n)
    {
      nums1.resize(m + n); // <= ENSURE SUFFICIENT STORAGE IN nums1
      std::merge(
        nums1.rbegin() + n, nums1.rend(), // consider that nums1 is already resized
        nums2.rbegin(), nums2.rend(),
        nums1.rbegin(),
        std::greater<int>());
    }
};

int main()
{
  // sample data
  std::vector<int> nums1{ 1, 3, 5, 7, 9 };
  std::vector<int> nums2{ 2, 4, 6, 8 };
  // run merge
  Solution().merge(nums1, (int)nums1.size(), nums2, (int)nums2.size());
  // output result
  std::cout << "nums1: {";
  const char* sep = " ";
  for (const int num : nums1) {
    std::cout << sep << num;
    sep = ", ";
  }
  std::cout << " }" << std::endl;
}

Output:

nums1: { 1, 2, 3, 4, 5, 6, 7, 8, 9 }

Live Demo on coliru

Notes:

1. I used rbegin() and rend() , to merge the vectors from end to begin, like OP did it. This is necessary to write merged elements always after where the still unprocessed elements have to be read.
2. The input iterators for nums1 may look a bit surprising on the 1 ^st glance. Using nums1.rbegin() + n is necessary to address to skip the space which was already allocated to store the additional elements of nums2 . (That are exactly n .)
3. The std::merge() has to be used with a custom predicate (eg std::greater ) as I used ascending ordered sample data but merge from back to front which reverses that into a descending order .