Interview Classic 150 Questions P88 Merge Two Sorted Arrays

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Timeline

timeline

2025-09-27

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two pointers

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Title:

P88 Merge Two Sorted Arrays

https://leetcode.cn/problems/merge-sorted-array/description/?envType=study-plan-v2&envId=top-interview-150

This problem can actually be merged in place, because the last n elements of num1 are 0, so you can merge from back to front. But the code below does not take this into account.
Note std::move, it actually converts the parameter t to an rvalue reference type, does not copy memory, does not release resources, does not call constructors or destructors

template <typename T>
typename std::remove_reference<T>::type&& move(T&& t) noexcept {
    return static_cast<typename std::remove_reference<T>::type&&>(t);
}

Take std::vectoras an example:

std::vector<int> a = {1, 2, 3};
std::vector<int> b = std::move(a);

Here: std::move(a) turns a into an rvalue reference. The compiler will choose the vector’s move constructor instead of the copy constructor.
What the move constructor does:

• Directly ‘steal’ the internal pointer of a and give it to b.
• Set a’s pointer to null to avoid releasing the same memory during destruction.

So in the end: b holds the data of {1,2,3}. a becomes empty (size()==0, but is still a valid object)

Summary:
nums1 = std::move(vec);

• Old content is automatically released (done internally by move assignment)
• New content takes over
• vec becomes empty
  No need to manually call the destructor.

#include <vector>
using std::vector;

class Solution {
    public:
        void merge(vector<int> &nums1, int m, vector<int> &nums2, int n)
        {
                vector<int> vec;
                int i = 0, j = 0;
                while (1) {
                        if (i < m && j < n) {
                                if (nums1[i] <= nums2[j])
                                        vec.push_back(nums1[i++]);
                                else
                                        vec.push_back(nums2[j++]);
                        } else if (i < m && j >= n) {
                                vec.push_back(nums1[i++]);
                        } else if (i >= m && j < n) {
                                vec.push_back(nums2[j++]);
                        } else {
                                break;
                        }
                }
                nums1 = std::move(vec);
        }
};