给出集合 [1,2,3,…,n]
,其所有元素共有 n! 种排列。
按大小顺序列出所有排列情况,并一一标记,当 n = 3 时, 所有排列如下:
"123"
"132"
"213"
"231"
"312"
"321"
给定 n 和 k,返回第 k 个排列。
说明:
- 给定 n 的范围是 [1, 9]。
- 给定 k 的范围是[1, n!]。
示例 1:
输入: n = 3, k = 3
输出: "213"
示例 2:
输入: n = 4, k = 9
输出: "2314"
#include "_000库函数.h" //一打眼看到题目描述,就想起用排列算法
//时间有点长588ms class Solution {
public:
string getPermutation(int n, int k) {
vector<int>v;
for (int i = ; i <= n; ++i)
v.push_back(i);
--k;
while (next_permutation(v.begin(), v.end()) && --k);
string s = "";
for (auto a : v)
s += a + '';
return s;
}
}; //用一下递归
class Solution {
public:
string getPermutation(int n, int k) {
vector<int>v;
for (int i = ; i <= n; ++i)
v.push_back(i);
Combin(v, , );
string s = "";
for (auto a : v)
s += a + '';
return s;
}
void Combin(vector<int>&v, int k, int s) {
if (!k)return;
if (s >= v.size())--k;
for (int i = s; i < v.size(); ++i) {
swap(v[i], v[s]);
Combin(v, k, s + );
swap(v[i], v[s]);
}
}
}; //博客答案,没怎么看懂
//后期看懂再更新
//12ms
class Solution {
public:
string getPermutation(int n, int k) {
string res;
string num = "";
vector<int> f(n, );
for (int i = ; i < n; ++i) f[i] = f[i - ] * i;
--k;
for (int i = n; i >= ; --i) {
int j = k / f[i - ];
k %= f[i - ];
res.push_back(num[j]);
num.erase(j, );
}
return res;
}
}; void T060() {
Solution s;
cout << s.getPermutation(, ) << endl;
cout << s.getPermutation(, ) << endl;
}