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 for n = 3: "123" "132" "213" "231" "312" "321&q…

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…

1. 原题链接 https://leetcode.com/problems/permutation-sequence/description/ 2. 题目要求 给出整数 n和 k ,k代表从1到n的整数所有排列序列中的第k个序列,返回String类型的第k个序列 3. 解题思路 首先我们要知道这个序列是按照什么规律排列下去的,假如此时n=4,k= 21,n=4时所有的排列如下: 可以看出 n=4 时,一共有 4!=24种排列组合. 每一个数字开头各有 6 种排列组合,因此我们可以把同一数字开头的…

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…

给出集合 [1,2,3,…,n],其所有元素共有 n! 种排列.按大小顺序列出所有排列情况,并一一标记,可得到如下序列 (例如,  n = 3):   1."123"   2. "132"   3. "213"   4. "231"   5. "312"   6. "321"给定 n 和 k,返回第 k 个排列序列.注意:n 介于1到9之间(包括9).详见:https://leetcod…

/* n个数有n!个排列,第k个排列,是以第(k-1)/(n-1)!个数开头的集合中第(k-1)%(n-1)!个数 */ public String getPermutation(int n, int k) { k--; List<Integer> list = new ArrayList<>(); StringBuilder res = new StringBuilder(); int count =1; //以每个数字开头的集合有多少中排列 for (int i = 2; i…

Given a linked list and a value x, partition it such that all nodes less than x come before nodes greater than or equal to x. You should preserve the original relative order of the nodes in each of the two partitions. For example, Given 1->4->3->…