![]() Note that some swaps are skipped in this example, since the input contains a duplicate 1 element: i=0, j=0 Here’s what happens when the input is 1,1,2. So it seems reasonable that this would cover every possible swap.Īs with last week’s recursive problem, instrumentation can help illuminate the flow of control in a recursive algorithm. So (ignoring duplicates), we swap position 0 with positions 0, 1, …, n-1, position 1 with 1, 2, …, n-1, and so on until i=n. And j covers every position from i to the last element. Notice that i covers every position in nums from 0 to the last element. The key is to make sure we use all possible swap positions (except where the swap would have no effect because the source and destination elements are the same). Why does this process work? Since permutations are arrangements of the input values, it makes sense that we could generate these arrangements by swapping. If we sent a reference, we would overwrite the result of previous swaps. This is how we generate new permutations. One implementation detail: When we make the recursive call, we need to send a copy of nums, not just a reference to nums. To keep the code simple, we also do the swap in this case, though it has no effect since we’re swapping an element with itself. As a special case, we also make the recursive call when i=j. If they’re different, we swap them and recursively start the process again at the next starting position i+1. For each j (the current position), we’ll check the values at nums and nums. The iteration loop on j goes from i (the starting position) to the end of the current permutation, nums. We’ll use a combination of iteration and recursion. If it is, the current permutation is done, so we can add it to our list of results, and return: if i is past the end of the current permutationĪdd the current permutation to the result
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