# 思路1 利用数学特征

首先n个数全排列 = n!

直接用回溯法做的话需要在回溯到第k个排列时终止就不会超时了, 但是效率依旧感人 可以用数学的方法来解, 因为数字都是从1开始的连续自然数, 排列出现的次序可以推 算出来, 对于n=4, k=15 找到k=15排列的过程:

1 + 对2,3,4的全排列 (3!个)\
2 + 对1,3,4的全排列 (3!个) 3, 1 + 对2,4的全排列(2!个) 3 + 对1,2,4的全排列 (3!个)-------> 3, 2 + 对1,4的全排列(2!个)-------> 3, 2, 1 + 对4的全排列(1!个)-------> 3214 4 + 对1,2,3的全排列 (3!个) 3, 4 + 对1,2的全排列(2!个) 3, 2, 4 + 对1的全排列(1!个)

确定第一位: k = 14(从0开始计数) index = k / (n-1)! = 2, 说明第15个数的第一位是3 1234 更新k k = k - index*(n-1)! = 2 确定第二位: k = 2 index = k / (n-2)! = 1, 说明第15个数的第二位是2 124 更新k k = k - index*(n-2)! = 0 确定第三位: k = 0 index = k / (n-3)! = 0, 说明第15个数的第三位是1 14 更新k k = k - index\*(n-3)! = 0 确定第四位: k = 0 index = k / (n-4)! = 0, 说明第15个数的第四位是4 4 最终确定n=4时第15个数为3214

数学原理参考`康拓展开`


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