Also, I wouldn't want to generate ALL permutations and return them, but only generating a single permutation, at a time, and continuing only if necessary ... ... <看更多>

public IEnumerable<int[]> Permut(int[] arr){. while (true){. yield return arr;. var j = arr.Length - 2;. while (j >= 0 && arr[j] >= arr[j+1]) j--;. ... <看更多>

I can easily generate permutations of a list with Permutations - but I feel like generating permutations and then looping over every element, ... ... <看更多>

The recursive overhead can be eliminated in a very simple way: instead of generating the sequence of permutations recursively, generate the ... ... <看更多>