[题目描述] Given n and k, return the k-th permutation sequence. Notice:n will be between 1 and 9 inclusive. 给定n和k,求123..n组成的排列中的第k个排列. [注]1 ≤ n ≤ 9 [题目链接] www.lintcode.com/en/problem/permutation-sequence/ [题目解析] 这道题给了我们n还有k,在数列 1,2,3,... , n构建的全排列中,返回第k个…

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): Given n and k, return the kth permutation sequence. Note: Given n will be between…

题目: 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" &qu…

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…

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…

一.开篇 既上一篇<交换法生成全排列及其应用> 后,这里讲的是基于全排列 (Permutation)本身的一些问题,包括:求下一个全排列(Next Permutation):求指定位置的全排列(Permutation Sequence):给出一个全排列,求其所在位置. 二.例题 1. 求下一个全排列,Next permuation Implement next permutation, which rearranges numbers into the lexicographically ne…

Next Permutation: Implement next permutation, which rearranges numbers into the lexicographically next greater permutation of numbers. If such arrangement is not possible, it must rearrange it as the lowest possible order (ie, sorted in ascending ord…