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grayCode.cpp
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grayCode.cpp
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// Source : https://oj.leetcode.com/problems/gray-code/
// Author : Hao Chen
// Date : 2014-06-20
/**********************************************************************************
*
* The gray code is a binary numeral system where two successive values differ in only one bit.
*
* Given a non-negative integer n representing the total number of bits in the code,
* print the sequence of gray code. A gray code sequence must begin with 0.
*
* For example, given n = 2, return [0,1,3,2]. Its gray code sequence is:
*
* 00 - 0
* 01 - 1
* 11 - 3
* 10 - 2
*
* Note:
* For a given n, a gray code sequence is not uniquely defined.
*
* For example, [0,2,3,1] is also a valid gray code sequence according to the above definition.
*
* For now, the judge is able to judge based on one instance of gray code sequence. Sorry about that.
*
**********************************************************************************/
#include <stdio.h>
#include <stdlib.h>
#include <iostream>
#include <vector>
using namespace std;
vector<int> grayCode(int n) {
vector<int> v;
//n = 1<<n;
int x =0;
v.push_back(x);
for(int i=0; i<n; i++){
int len = v.size();
for (int j=0; j<len; j++){
x = v[j]<<1;
if (j%2==0){
v.push_back(x);
v.push_back(x+1);
}else{
v.push_back(x+1);
v.push_back(x);
}
}
v.erase(v.begin(), v.begin()+len);
}
return v;
}
void printBits(int n, int len){
for(int i=len-1; i>=0; i--) {
if (n & (1<<i)) {
printf("1");
}else{
printf("0");
}
}
}
void printVector(vector<int>& v, int bit_len)
{
vector<int>::iterator it;
for(it=v.begin(); it!=v.end(); ++it){
//bitset<bit_len> bin(*it);
printBits(*it, bit_len);
cout << " ";
//cout << *it << " ";
}
cout << endl;
}
int main(int argc, char** argv)
{
int n = 2;
if (argc>1){
n = atoi(argv[1]);
}
vector<int> v = grayCode(n);
printVector(v, n);
return 0;
}