Leetcode算法学习日志-60 Permutation Sequence

Leetcode 60 Permutation Sequence

题目原文

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.

题意分析

寻找按字典排序的第k个序列。

解法分析

对于一个n，可以构建n的所有排列的排列树，树的叶子数代表总的排列数，共n！个，对于第一层的n个子树，每个子树包含(n-1)!个序列，通过k与(n-1）!的取商取余，可以逐步寻找到相应解，c++代码如下：

class Solution {
private:
int N;//total path in each layer
int curPath;
int K;//the order in each layer

public:
int pathNumber(int layer) {
int i;
curPath = 1;
for (i = 1; i <= (N - layer); i++) {
curPath = curPath*i;
}
int temp = K / (curPath / (N - layer));
if (K % (curPath / (N - layer)) == 0) {
K = curPath / (N - layer);
return temp - 1;
}
else {
K = K % (curPath / (N - layer));
return temp;
}

}
string getPermutation(int n, int k) {
N = n;
K = k ;
string res;
vector<int> num;
int i;
for (i = 1; i <= n; i++) {
num.push_back(i);
}
if (n == 1)
return "1";
int whichWay;
for (i = 0; i<n; i++) {
whichWay = pathNumber(i);
res.push_back('0'+num[whichWay]);
num.erase(num.begin() + whichWay);
}
//res.push_back(num[0]);
return res;
}
};