LeetCode OJ--Permutation Sequence *】的更多相关文章

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 (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" "…
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…