Permutation Sequence 求第k个的排列序列 @LeetCode

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思路：

1 求出所有的排序，直到k为止。至少Java会超时。

2 数学，找规律，不好想！参考 http://fisherlei.blogspot.com/2013/04/leetcode-permutation-sequence-solution.html

假设有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个元素共有(n-1)!组排列，那么这里就可以知道

设变量K1 = K
a1 = K1 / (n-1)!// 第一位的选择下标

同理，a2的值可以推导为

K2 = K1 % (n-1)!
a2 = K2 / (n-2)!

。。。。。

K(n-1) = K(n-2) /2!
a(n-1) = K(n-1) / 1!

an = K(n-1)

package Level5;import java.util.Arrays;/**Permutation SequenceThe 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""321"Given n and k, return the kth permutation sequence.Note: Given n will be between 1 and 9 inclusive. * */public class S60 {public static void main(String[] args) {System.out.println(getPermutation(3, 3));}static int maxDep;static StringBuilder ret;static boolean[] canUse;static int[] num;static int kth;static String save = "";// 数学方法public static String getPermutation(int n, int k) {int[] nums = new int[n+10];int permCount = 1;for(int i=0; i<n; i++){nums[i] = i+1;// nums装有1,2,3,4,...,npermCount *= (i+1);// 最后计算出permCount = n!}k--;// 对k减一，因为现在index是从[0,n-1]而不是[1,n]StringBuilder sb = new StringBuilder();for(int i=0; i<n; i++){permCount = permCount / (n-i);int idx = k / permCount;// 该位应该选择的下标sb.append(nums[idx]);// 重建nums，左移一位for(int j=idx; j<n-i; j++){nums[j] = nums[j+1];}k %= permCount;}return sb.toString();}// 递归求所有的排列，超时public static String getPermutation2(int n, int k) {        kth = k;StringBuilder done = new StringBuilder();StringBuilder rest = new StringBuilder();for(int i=1; i<=n; i++){rest.append(i);}rec(done, rest);return save;    }public static void rec(StringBuilder done, StringBuilder rest){if(rest.length() == 0){kth--;if(kth == 0){save = done.toString();}return;}for(int i=0; i<rest.length(); i++){char c = rest.charAt(i);rest.deleteCharAt(i);done.append(c);rec(done, rest);done.deleteCharAt(done.length()-1);rest.insert(i, c);}}}

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