原题链接: http://oj.leetcode.com/problems/permutation-sequence/ 这道题目算法上没有什么特别的,更像是一道找规律的数学题目。我们知道,n个数的permutation总共有n阶乘个,基于这个性质我们可以得到某一位对应的数字是哪一个。思路是这样的,比如当前长度是n,我们知道每个相同的起始元素对应(n-1)!个permutation,...
Can you solve this real interview question? 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 for n = 3: 1. "123" 2. "1
This is a dynamic programming question. Usually, solving and fully understanding a dynamic programming problem is a 4 step process: Start with the recursive backtracking solution (从递归回溯法入手) Optimize by using a memoization(记忆) table (top-down[3] dynamic programming) Remove the need for ...
The set [1,2,3,...,n] contains a total of n! unique permutations. By listing and labeling all of thepermutationsin order, we get the following sequence for n = 3: "123" "132" "213" "231" "312" "321" Given n and k, return the kth permutation sequence. Note: Given n will ...
【LeetCode】#60第k个排列(Permutation Sequence) 题目描述 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,当 n = 3 时, 所有排列如下: “123” “132” “213” “231” &ld...Leetcode60. Permutation Sequence第k个排列 给出集合 [1,2,3,…,...
Givennandk, return thekthpermutation sequence. 如果有n个元素,第K个permutation是a1, a2, a3, ... ..., an,那么a1是哪一个数字呢? 那么这里,我们把a1去掉,那么剩下的permutation为:a2, a3, ... ... an, 共计n-1个元素。 n-1个元素共同拥有(n-1)!组排列,那么这里就能够知道 设变量...
http:///2013/04/leetcode-permutation-sequence-solution.html 思路: 分别计算第k个permutation的各个字母。 AI检测代码解析 class Solution(object): def getPermutation(self, n, k): """ :type n: int :type k: int :rtype: str """ res = '' ...
LeetCode --- 60. Permutation Sequence 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):...
LeetCode Permutation Sequence Permutation Sequence The set[1,2,3,…,n]contains a total ofn! unique permutations. By listing and labeling all of the permutations in order, We get the following sequence (ie, forn= 3): "123" "132" "213"...
因为题目要求(Leetcode 31),这里需要重新变成顺序(要求再次变成1 2 3 4 5)。然而在后面的两题中,以及C++STL的next_permutation实现中,都没有必要。 class Solution { public: void nextPermutation(vector<int>& nums) { int i,j; for (i=nums.size()-1; i>0 && nums[i-1] >= nums[i]; i--)...