/* 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…
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…
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…
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…
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…
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…