在 C 中，为什么将 x 与 y 交换比将 y 与 x 交换更快？

Question

In a problem minimum-swaps-2 , arr is unsorted array, arr_count is number of array elements, we need to return minimum number of swaps to sort the array,

i tried:

int minimumSwaps(int arr_count, int* arr) {
    int swap = 0;
    for(int i  = 0; i < arr_count;){
        if(arr[i] != (i + 1)){
            int temp = arr[i];
            arr[i] = arr[arr[i] - 1];
            arr[arr[i] - 1] = temp;
            swap++;
        } else {
            i++;
        }
    }
    return swap;
}

but it didn't work. It shows time out error. ie it took more time to run! Then i tried the below code, it worked!

        int temp = arr[arr[i] - 1];
        arr[arr[i] - 1] = arr[i];
        arr[i] = temp;
        swap++;

The only difference between both is swapping x with y or y with x. What difference does it make?

Answer 1

After arr[i] = arr[arr[i] - 1]; the value arr[i] was changed and arr[arr[i] - 1] doesn't point at same storage anymore, arr[arr[i] - 1] = temp; writes to wrong storage.

So you actually do not swap two values. It could be obvious if you represented it as pointer arithmetic.

        int temp = *(arr + i);
        *(arr + i) = *(arr + temp - 1);
        *(arr + *(arr + i) - 1) = temp;

Last line equals to *(arr + *(arr + temp - 1) - 1) with values present at moment before line 2 execution, which IS wrong by definition of task.

In fact it gives obvious solution:

        int temp = *(arr + i);
        *(arr + i) = *(arr + temp - 1);
        *(arr + temp - 1) = temp;

or

        int temp = arr[i];
        arr[i] = arr[temp - 1];
        arr[temp - 1] = temp;

Which yields result of 5 returned from minimumSwaps .

Essentially you time-outed because first variant unable to reach solution.