LeetCode 89: Gray Code解题报告

89. Gray Code


提交网址:  https://leetcode.com/problems/gray-code/ 

Total Accepted: 58554   Total Submissions: 161869   Difficulty: Medium   ACrate:  36.2%

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.


分析:
  格雷码 有个相应的数学公式,整数n的格雷码是n^(n/2).

    方法1: 最简单的方法,利用数学公式,对从0~2^n-1的所有整数,转化为格雷码。
    方法2: n比特的格雷码,可以递归地从n-1比特的格雷码生成。如下图所示:

LeetCode 89: Gray Code解题报告_第1张图片


Gray Code 0 = 0, 下一项是toggle最右边的bit(LSB), 再下一项是toggle最右边值为 “1” bit的左边一个bit。然后重复

如: 3bit

Gray Code:  000, 001, 011, 010110, 111, 101, 100, 最右边值为 “1” 的bit在最左边了,结束。

Binary      :  000, 001, 010, 011, 100, 101, 110, 111

 

再者就是Binary Code 转换为Gray Code了。

如:

  Binary Code :1011 要转换成Gray Code

  1011 = 1(照写第一位), 1(第一位与第二位异或 1^0 = 1), 1(第二位异或第三位, 0^1=1), 0 (1^1 =0) = 1110

  其实就等于 (1011 >> 1) ^ 1011 = 1110


AC代码:

#include<iostream>
#include<vector>
using namespace std;
 
class Solution {
public:
    vector<int> grayCode(int n) {
        vector<int> emptyArr(0);
        if(n<0) return emptyArr;    // 输入异常时,即n为负整数时的处理办法...
        if(n>=0)
        {        
        int size=1<<n;
        vector<int> NumArr(size,0);      
          
        int gNum=0;
        for(int i=0;i<size;i++)
        {
            gNum=i ^ i>>1;    // 数学公式G[i]=i ^ i/2,gNum是一个格雷码对应的二进制数,存储时变成int(十进制)
            NumArr[i]=gNum;
        }
        return NumArr;
        }
    }
};
 
/*  以下为测试 */
/*
int main()
{
    int n,j;
    vector<int> display;
    cin>>n;
     
    Solution sol;
    display=sol.grayCode(n);
    for(j=0;j<display.size();j++)
    {
        cout<<display[j]<<" ";
    }
    cout<<endl;
    return 0;
}*/

腾讯2016研发工程师编程题  考过这个题:
http://www.nowcoder.com/questionTerminal/50959b5325c94079a391538c04267e15

相关链接:

2016校招腾讯研发岗笔试题(第一题 格雷码) - u010660138的博客 - CSDN http://blog.csdn.net/u010660138/article/details/48265341


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