请参阅QuickPerm算法，它是迭代的：http ://www.quickperm.org/
编辑：
为了清楚起见，在Ruby中进行了重写：
def permute_map(n) results = [] a, p = (0...n).to_a, [0] * n i, j = 0, 0 i = 1 results << yield(a) while i < n if p[i] < i j = i % 2 * p[i] # If i is odd, then j = p[i], else j = 0 a[j], a[i] = a[i], a[j] # Swap results << yield(a) p[i] += 1 i = 1 else p[i] = 0 i += 1 end end return resultsend
