LeetCode || Permutation Sequence
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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"
"321"
Given n and k, return the kth permutation sequence.
Note: Given n will be between 1 and 9 inclusive.
这个题折磨了好几天,不能按顺序一个一个的求了,那样时间复杂度太高了
后来在这个博客里看到了方法:
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)
class Solution {public: string getPermutation(int n, int k) {string permSueq;if(n>9)return permSueq;int factor=getFactorial(n);vector<char> char_v;for(int i=1;i<=n;++i){char_v.push_back(getChar(i));}--k;//别忘了首先减一下,由于数组索引是从0开始的while(n>0){factor/=n;int pos=(k/factor)%char_v.size();vector<char>::iterator iter=char_v.begin()+pos;permSueq.push_back(*iter);char_v.erase(iter);//删除已经使用了的数字,这样不会影响下次循环k%=factor;--n;}return permSueq;}private://阶乘int getFactorial(int n){if(n==1||n==0)return 1;elsereturn n*getFactorial(n-1);}char getChar(int n){switch(n){case 1:return '1';case 2:return '2';case 3:return '3';case 4:return '4';case 5:return '5';case 6:return '6';case 7:return '7';case 8:return '8';case 9:return '9';default:return 'a';}}};
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