标签：style blog http io ar color sp for strong
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""312""321"
Given n and k, return the kth permutation sequence.
Note: Given n will be between 1 and 9 inclusive.
我的方法从高位到低位逐位确定数字。
以图例n=3进行说明：
构建数组v={1,2,3}
确定最高位：
ind=(k-1)/2
注：分母2是指每个最高位持续的排列数。由于除了最高位之外还有n-1=2位，一共可以出现2!种排列。
ind指的是所求的第k个排列会落在哪一个最高位的覆盖范围内。
k==1,2时，ind==0，最高位为v[ind]==1
k==3,4时，ind==1，最高位为v[ind]==2
k==5,6时，ind==2，最高位为v[ind]==3
其余数位如上思路逐位确定。注意k的取余更新。
class Solution { public: string getPermutation(int n, int k) { vector<int> v(n, 0); for(int i = 0; i < n; i ++) v[i] = i+1; string result = ""; while(n-1) { int divisor = fac(n-1); int ind = (k-1)/divisor; //itoa string digit; char temp[32]; sprintf(temp, "%d", v[ind]); digit = temp; result += digit; for(int i = ind+1; i < v.size(); i ++) v[i-1] = v[i]; v.pop_back(); k %= divisor; if(k == 0) k = divisor; n --; } //itoa string digit; char temp[32]; sprintf(temp, "%d", v[0]); digit = temp; result += digit; return result; } int fac(int n) { if(n == 0) return 1; int result = 1; for(int i = 1; i <= n; i ++) result *= i; return result; } };
【LeetCode】Permutation Sequence
标签：style blog http io ar color sp for strong
原文地址：http://www.cnblogs.com/ganganloveu/p/4156188.html
