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Copy pathpermutationSequence.cpp
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168 lines (140 loc) · 3.49 KB
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// Source : https://oj.leetcode.com/problems/permutation-sequence/
// Author : Hao Chen
// Date : 2014-08-22
/**********************************************************************************
*
* 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 kth permutation sequence.
*
* Note: Given n will be between 1 and 9 inclusive.
*
**********************************************************************************/
#include <stdlib.h>
#include <iostream>
#include <sstream>
#include <string>
#include <vector>
#include <algorithm>
using namespace std;
/*
"123"
"132"
"213"
"231"
"312"
"321"
*/
void nextPermutation(vector<int>& num);
/* Extreamly Optimized */
string getPermutation(int n, int k) {
vector<int> num;
int total = 1;
for(int i=1; i<=n; i++){
num.push_back(i);
total *= i;
}
//invalid k;
if( total < k ) {
return "";
}
// Construct the k-th permutation with a list of n numbers
// Idea: group all permutations according to their first number (so n groups, each of
// (n-1)! numbers), find the group where the k-th permutation belongs, remove the common
// first number from the list and append it to the resulting string, and iteratively
// construct the (((k-1)%(n-1)!)+1)-th permutation with the remaining n-1 numbers
int group = total;
stringstream ss;
while (n>0) {
group = group / n;
int idx = (k-1) / group;
ss << num[idx];
num.erase(num.begin()+idx);
n--;
//the next k also can be caculated like this:
// k = (k-1)%group + 1;
k -= group * idx;
}
return ss.str();
}
/* Optimization by determining the group */
string getPermutation_0(int n, int k) {
vector<int> num;
int total = 1;
for(int i=1; i<=n; i++){
num.push_back(i);
total *= i;
}
//invalid k;
if( total < k ) {
return "";
}
int group = total / n;
int idx = (k-1) / group;
int nn = num[idx];
num.erase(num.begin()+idx);
num.insert(num.begin(), nn);
int offset = (k-1) % group;
for(int i=0; i<offset; i++) {
nextPermutation(num);
}
//string result;
stringstream ss;
for(int i=0; i<n; i++){
ss << num[i];
}
return ss.str();
}
/* Time Limit Exceeded */
string getPermutation_1(int n, int k) {
vector<int> num;
for(int i=1; i<=n; i++){
num.push_back(i);
}
for(int i=1; i<k; i++) {
nextPermutation(num);
}
//string result;
stringstream ss;
for(int i=0; i<n; i++){
ss << num[i];
}
return ss.str();
}
void nextPermutation(vector<int>& num) {
if (num.size()<=1) return;
for (int i=num.size()-1; i>0; i-- ) {
if (num[i-1] < num[i]) {
int j = num.size() - 1;
while( num[i-1] > num[j] ) {
j--;
}
int temp = num[i-1];
num[i-1] = num[j];
num[j] = temp;
reverse(num.begin()+i, num.end());
return;
}
}
reverse( num.begin(), num.end() );
}
int main(int argc, char**argv)
{
int n=3, k=6;
if ( argc > 2 ) {
n = atoi(argv[1]);
k = atoi(argv[2]);
}
cout << "n = " << n << ", k = " << k << " : " << getPermutation(n, k) << endl;
return 0;
}