60. Permutation Sequence

60. Permutation Sequence

• Total Accepted: 65886
• Total Submissions: 249649
• Difficulty: Medium

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 k^th permutation sequence.

刚开始用全排列的常规方法，DFS进行顺序查找，超时

public class Solution {    static int index = 0;    public String getPermutation(int n, int k) {        int[] visited = new int[n];        index = 0;        return permutate(visited, n, k, "");    }        String permutate(int[] visited, int n, int k, String nowStr){               String result = "";        if(nowStr.length() == n){            index++;           return nowStr;        }else{            for(int i = 1; i <= n; i++){                if(visited[i-1] == 0){                    visited[i-1] = 1;                    nowStr += i;                    result = permutate(visited, n, k, nowStr);                    //System.out.println(index);                    if(index == k){                        return result;                    }                    nowStr = nowStr.substring(0, nowStr.length()-1);                    visited[i-1] = 0;                }            }        }                return result;    }    }

转换思路，An, An-1, ..... , A2, A1   An能表示的数字有n！，利用此规律，第一次确定最高位，第二次确定次高位。。。。。。。循环即可得到结果

public class Solution {    static int num = 0;    public String getPermutation(int n, int k) {        String result = "";        int[] nums = new int[n+1];        int[] visited = new int[n+1];        nums[1] = 1;        for(int i = 2; i <= n; i ++){            nums[i] = i * nums[i-1];        }                for(int i = n; i >= 2; i --){            int times = k / nums[i-1];            if( k % nums[i-1] != 0){                times++;                k = k % nums[i-1];            }else{                k = nums[i-1];            }            int item = 1;            for(int j = 1; j <= n; j++){                if(visited[j] == 0){                    if(times >= 0){                        times--;                    }                    if(times == 0){                        result += j;                        visited[j] = 1;                        break;                    }                }            }                    }        for(int j = 1; j <= n; j++){            if(visited[j] == 0){                result += j;                visited[j] = 1;            }        }        return result;    }      }