leetcode_89题——Gray Code（回溯法）

Gray Code

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Question Solution

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.

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Backtracking

这道题的意思是给出n值时，给出n位的格雷码所对应的数字的向量，这道题采用回溯法，因为n位的与n-1位的格雷码是有关系的，前半部分与n-1位的相同。后半部分的由对称的前面那部分加上2^n-1，所以采用递归的往回溯。

下面是AC的代码

#include<iostream>
#include<vector>
#include <queue>
using namespace std;

int conversion(int n)
{
	int temp=1;
	if(n==1)
		return 1;
	for(int i=1;i<n;i++)
		temp=temp*2;
	return temp;
}
vector<int> grayCode(int n) {
	vector<int> result;
	if(n==0)
	{
		result.push_back(0);
		return result;
	}
	if(n==1)
	{
		result.push_back(0);
		result.push_back(1);
		return result;
	}
	else
	{
		//queue<int> temp;
		result=grayCode(n-1);
		int len=result.size();
		for(int i=len-1;i>=0;--i)
		{
			int temp=result[i]+conversion(n);
			result.push_back(temp);
		}
		return result;
	}
}

int main()
{
	vector<int> temp=grayCode(4);

	for(int i=0;i<temp.size();i++)
		cout<<temp[i]<<endl;
}