[Leetcode]Permutation Sequence

最新推荐文章于 2019-05-17 11:20:47 发布

原创最新推荐文章于 2019-05-17 11:20:47 发布 · 273 阅读

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本文探讨了如何通过递归和迭代方法解决排列序列问题，具体展示了使用递归实现深度优先搜索（DFS）求解排列组合，并提供了一个改进的算法来减少时间复杂度。同时，还介绍了一种基于数学原理的快速查找特定排列的方法，适用于给定n和k的情况下直接获取第k个排列序列。

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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 k^th permutation sequence.

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

class Solution {
public:
    /*algorithm dfs  TLE error
   */
    void dfs(int n,int k,string &result,int &count,vector<int>&path,vector<int>&accessed){
        if(path.size() == n){
            ++count;
            if(count == k){ //find kth
                for(int i = 0;i < n;i++)
                    result.append(1,path[i] + '0');
            }
            return;
        }
        for(int i = 1;i <= n;i++){
            if(accessed[i])continue;
            accessed[i] = 1;
            path.push_back(i);
            dfs(n,k,result,count,path,accessed);
            path.pop_back();
            accessed[i] = 0;
            if(count == k)return;
        }
    }
    string getPermutation(int n, int k) {
        string result;
        vector<int>path;
        vector<int>visited(n,0);
        int count = 0;
        dfs(n,k,result,count,path,visited);
        return result;
    }
};

class Solution {
public:
    /*algorithm 
      n!=(n-1)*n
      A[]={1,...,n}
      let t:(n-1)!
      1)index = (k-1)/t,  r=(k-1)%t + 1
      2)A remove A[index]
      3)k = r,t = t/A.size()
      time O(n) ,space O(n)
    */
    string getPermutation(int n, int k) {
        string s;
        vector<int>nums(n,0);
        int t = 1; //(n-1)!
        for(int i = 1;i <= n;i++){
            nums[i-1]=i;
            if(i < n)t *= i;
        }
        while(n){
            int i = (k-1)/t;
            k = (k-1)%t + 1;
            s.append(1,nums[i]+'0');
            nums.erase(nums.begin()+i);
            n--;
            if(n)t /= n;
        }
        return s;
    }
};