zoukankan      html  css  js  c++  java
  • 【LeetCode】060. Permutation Sequence

    题目:

    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):

    1. "123"
    2. "132"
    3. "213"
    4. "231"
    5. "312"
    6. "321"

    Given n and k, return the kth permutation sequence.

    Note: Given n will be between 1 and 9 inclusive.

    题解:

    Solution 1 

    class Solution {
    public:
        string getPermutation(int n, int k) {
            string s;
            for(int i = 0; i < n; ++i){
                s += (i + 1) + '0';
            }
            for(int i = 0; i < k - 1; ++i){
                next_permutation(s);
            }
            return s;
        }
        void next_permutation(string &str){
            int n = str.size();
            for(int i = n - 2; i >= 0; --i){
                if(str[i] >= str[i + 1]) continue;
                int j = n - 1;
                for(; j > i; --j) {
                    if(str[j] > str[i]) break;       
                }
                swap(str[i], str[j]);
                reverse(str.begin() + i + 1, str.end());
                return;
            }
            reverse(str.begin(), str.end());
        }
    };

    Solution 2 

    class Solution {
    public:
        string getPermutation(int n, int k) {
            string res;
            if(n <= 0 || k <= 0){
                return res;
            }
            string num = "123456789";
            vector<int> f(n, 1);
            for(int i = 1; i < n; ++i){
                f[i] = f[i - 1] * i;
            }
            --k;
            for(int i = n; i > 0; --i){
                int j = k / f[i - 1];
                k %= f[i - 1];
                res.push_back(num[j]);
                num.erase(j, 1);
            }
            return res;
        }
    };

    康托编码

    Solution 3

    class Solution {
    public:
        string getPermutation(int n, int k) {
            string s = "123456789", str;
            int factorial = 1;
            for(int i = 1; i < n; ++i){
                factorial *= i;
            }
            --k;
            for(int i = n; i > 0; --i){
                int index = k / factorial;
                str += s[index];
                s.erase(index, 1);
                k %= factorial;
                factorial /= i - 1 ? i - 1 : 1;
            }
            return str;
        }
    };
  • 相关阅读:
    在ASP.Net和IIS中删除不必要的HTTP响应头
    Json对象与Json字符串互转
    Jquery ajax传递复杂参数给WebService
    HTTP的KeepAlive是开启还是关闭?
    MQ产品比较-ActiveMQ-RocketMQ
    RocketMQ(7)——通信协议
    mq使用经验
    mq
    RocketMQ
    发送短信验证码实现方案
  • 原文地址:https://www.cnblogs.com/Atanisi/p/7477303.html
Copyright © 2011-2022 走看看