leetcode || 89、Gray Code

problem:

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

题意：首先要知道格雷码的定义，求N位的格雷码，并把它转换成 int 型输出

thinking:

（1）严格按照格雷码定义来，格雷码有一个递推公式：

利用上述公式可以递推（回溯）出N位格雷码，我这里采用字符串的处理方式实现上述公式，原理一样。

code:

class Solution {
public:
    vector<int> grayCode(int n) {
        vector<int> ret;
        if(n==0)
        {
            ret.push_back(0);
            return ret;
        }
        if(n==1)
        {
            ret.push_back(0);
            ret.push_back(1);
            return ret;
        }
        if(n==2)
        {
            ret.push_back(0);
            ret.push_back(1);
            ret.push_back(3);
            ret.push_back(2);
            return ret;
        }
        vector<string> gc2;
        gc2.push_back(string("00"));
        gc2.push_back(string("01"));
        gc2.push_back(string("11"));
        gc2.push_back(string("10"));
        vector<string>  gc(gc2);
        int length=4;

        for(int i=3;i<=n;i++)
        {
            length*=2;
            vector<string> gc_tmp(length,string(""));
            for(int j=0;j<length/2;j++)
            {
                string str0="0"+gc[j];
                string str1="1"+gc[length/2-j-1];
                gc_tmp[j]=str0;
                gc_tmp[j+length/2]=str1;
            }
            gc=gc_tmp;

        }
        for(int i=0;i<gc.size();i++)
            ret.push_back(string_to_int(gc[i]));
        return ret;

    }
protected:
    int string_to_int(string str)
    {
        cout<<str<<endl;
        int m=str.size();
        double res=0;
        for(int i=0;i<m;i++)
        {
            res+=(str.at(i)-'0')*pow(2,m-i-1);
        }
        return int(res);
    }
};