题目大意:给出n和k,找到1..n这些数组成的有序全排列中的第k个. 首先,n的全排列可以分成n组,每一组由n-1个数组成. 例如  3的全排列,分成三组: 1 2 3  和 1 3 2 2 1 3  和 2 3 1 3 1 2  和 3 2 1 每一组的个数是(n-1)!,每一组的打头的都是i .第i组以i打头. 先求 x = (k-1) / (n-1)!  + 1 比如  n = 3, k = 4. x = 3/2 + 1 = 2,得到 第四个实际上第二组里面. y = (k-1)%(n-1…

LeetCode:60. Permutation Sequence,n全排列的第k个子列 : 题目: LeetCode:60. Permutation Sequence 描述: The set [1,2,3,-,n] contains a total of n! unique permutations. By listing and labeling all of the permutations in order, We get the following sequence (ie, for…

The set [1,2,3,…,n] contains a total of n! unique permutations. By listing and labeling all of the permutations in order, We get the following sequence (ie, for n = 3): "123" "132" "213" "231" "312" "…

The set [1,2,3,…,n] contains a total of n! unique permutations. By listing and labeling all of the permutations in order,We get the following sequence (ie, for n = 3): "123" "132" "213" "231" "312" "3…

The set [1,2,3,…,n] contains a total of n! unique permutations. By listing and labeling all of the permutations in order,We get the following sequence (ie, for n = 3): "123" "132" "213" "231" "312" "3…

The set [1,2,3,-,n] contains a total of n! unique permutations. By listing and labeling all of the permutations in order,We get the following sequence (ie, for n = 3): "123" "132" "213" "231" "312" "3…

The set [1,2,3,…,n] contains a total of n! unique permutations. By listing and labeling all of the permutations in order,We get the following sequence (ie, for n = 3): "123" "132" "213" "231" "312" "3…

描述: The set [1,2,3,…,n] contains a total of n! unique permutations. By listing and labeling all of the permutations in order, We get the following sequence (ie, for n = 3): "123" "132" "213" "231" "312" &q…

Permutation Sequence The set [1,2,3,-,n] contains a total of n! unique permutations. By listing and labeling all of the permutations in order, We get the following sequence (ie, for n = 3): "123" "132" "213" "231" &…

The set [1,2,3,…,n] contains a total of n! unique permutations. By listing and labeling all of the permutations in order,We get the following sequence (ie, for n = 3): "123" "132" "213" "231" "312" "3…