C++版 - 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:

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, 010, 110, 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;
}*/