Leetcode Permutation Sequence

超时算法—排序dfs：

class Solution {
public:
string s;
string s1;
int count;string getPermutation(int n, int k) {s="123456789";count=0;s1="";dfs(0,n,k);return s1;}void dfs(int depth,int n,int k){if(depth==n&&count<=k){count++;if(count==k){s1=s.substr(0,n);return;}}if(depth<n){dfs(depth+1,n,k);for(int i=depth+1;i<n;i++){string s2=s;swap(s[depth],s[i]);sort(s.begin()+depth+1,s.begin()+n);dfs(depth+1,n,k);s=s2;}}}
};

超时算法—组合dfs

class Solution {
public:
string s1;
bool used[9];
int count;
char c[9];string getPermutation(int n, int k) {for(int i=0;i<9;i++){used[i]=false;c[i]='0';}        count=0;s1="";dfs(0,n,k);return s1;}void dfs(int depth,int n,int k){if(depth==n&&count<k){count++;if(count==k){for(int i=0;i<n;i++)s1+=c[i];return;}}if(depth<n){for(int i=0;i<n;i++){if(!used[i]){used[i]=true;c[depth]='0'+i+1;dfs(depth+1,n,k);used[i]=false;}}           }}
};

纯数学算法

class Solution {
public:string getPermutation(int n, int k) {string      result;vector<int> num(n);int         i, d(1), q;for(i=1;i<=n;++i){num[i-1]  = i;d        *= i;  //d=n!}for(i=n;i>=1;--i){d       = d/i;q       = (k-1)/d;k       = k-q*d;result.push_back('0'+num[q]);num.erase(num.begin()+q);}return result;}
};

转载于:https://www.cnblogs.com/tgkx1054/archive/2013/06/02/3114426.html

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