leetcode89-Gray Code(格雷码)

问题描述：

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.

问题求解：

n+1位格雷码一共有2^(n+1) = 2^n * 2个码字，可分为前2^n个和后2^n个。

规律如下：

1、(n+1)位格雷码（grayCode(n+1)）中的前2^n个码字等于：n位格雷码的
码字（grayCode(n)）按顺序书写，加前缀0。

2、(n+1)位格雷码（grayCode(n+1)）中的后2^n个码字等于：n位格雷码的
码字（grayCode(n)）按逆序书写，加前缀1。

由于是二进制，在最高位加0跟原来的数本质没有改变，所以取得上一位算出的格雷码结果，再加上逆序加1的结果就是当前这位格雷码的结果。

例子：

n = 1时，[0,1]->[0,1]
n = 2时，[00,01,11,10]->[0,1,3,2]
n = 3时，[ 000,001,011,010 , 110,111,101,100]->[0,1,3,2,6,7,5,4]

当n=1时，0，1

当n=2时，原来的0,1不变，只是前面形式上加了个0变成00，01。然后加数是1<<1为10，依次：10+1=11 10+0=10。结果为：00 01 11 10

当n=3时，原来的00,01,11, 10（倒序为：10，11，01，00） 。加数1<<2为100。倒序相加为：100+10=110, 100+11= 111 ,100+01= 101 , 100+00= 100。结果为000 001 011 010 110 111 101 100.十进制大小即是：0,1,3,2,6,7,5,4。

递归求解

class Solution {
public:
    vector<int> grayCode(int n) {
        if(n==0)
        {
            vector<int> res;
            res.push_back(0);
            return res;
        }
        vector<int> pre = grayCode(n-1);
        //(1)先把n-1位的格雷码放进结果数组(因为二进制在最高位加0与否没有影响)
        vector<int> res(pre);
        int npre = pre.size();
        for(int i=npre-1;i>=0;i--)
        {//(2)将把n-1位的格雷码逆序再与1<<n-1相加，最后放到结果数组
            res.push_back((1<<n-1) + pre[i]);
        }
        return res;
    }
};

当n=3时，输出结果：[0,1,3,2,6,7,5,4]

字符串形式：

#include<iostream>
#include<vector>
#include <string>
using namespace std;
class GrayCode {
public:
    vector<string> getGray(int n) {
        if(n==1)
        {
            vector<string> res;
            res.push_back("0");
            res.push_back("1");
            return res;
        }
        vector<string> pre = getGray(n-1);
        int npre = pre.size();
        int ncur = npre * 2;
        vector<string> res(ncur);
        for(int i=0;i<npre;i++)
        {
            res[i] = "0" + pre[i];//前2^(n-1)个
            res[ncur-i-1] = "1" + pre[i];//后2^(n-1)个
        }
        return res;
    }
};
int main()
{
    GrayCode g;
    vector<string> res = g.getGray(3);
    for(unsigned i=0;i<res.size();i++)
        cout<<res[i]<<endl;
    return 0;
}

结果：

000
001
011
010
110
111
101
100

Process returned 0 (0x0)   execution time : 0.591 s
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