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" "…

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…

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…

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…

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" "…

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…

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…

问题: 对于给定序列1...n,permutations共同拥有 n!个,那么随意给定k,返回第k个permutation.0 < n < 10. 分析: 这个问题要是从最小開始直接到k,预计会超时,受10进制转换为二进制的启示,对于排列,比方 1,2,3 是第一个,那么3!= 6,所以第6个就是3,2,1.也就是说,从開始的最小的序列開始,到最大的序列,就是序列个数的阶乘数.那么在1,3 , 2的时候呢?调整一下,变成2,1,3,就能够继续. 实现: int getFactorial(int…

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…

/* 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…

LeetCode 31 Next Permutation / 60 Permutation Sequence [Permutation] <c++> LeetCode 31 Next Permutation 给出一个序列,求其下一个排列 STL中有std::next_permutation这个方法可以直接拿来用 也可以写一个实现程序: 从右往左遍历序列,找到第一个nums[i-1]<num[i]的位置,记p = i-1. 如果第一步没有找到,说明整个序列满足单调递减,也就是最大的排列,那…

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" &q…

原题地址:https://oj.leetcode.com/submissions/detail/5341904/ 题意: 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&…

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…

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" &q…

一天一道LeetCode系列 (一)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): 1:"123" 2:"132"  3 : "213" 4 :&quo…

Permutation Sequence https://oj.leetcode.com/problems/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&…

Given an unsorted array of integers, find the length of the longest consecutive elements sequence. For example, Given [100, 4, 200, 1, 3, 2], The longest consecutive elements sequence is [1, 2, 3, 4]. Return its length: 4. Your algorithm should run i…

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&…

作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 目录 题目描述 题目大意 解题方法 日期 题目地址:https://leetcode.com/problems/permutation-sequence/description/ 题目描述 The set [1,2,3,...,n] contains a total of n! unique permutations. By listing and labeling all of th…

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…

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…

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…