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Saturday, December 22, 2012

[LeetCode] Gray Code 解题报告


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.
» Solve this problem


[解题思路]
看到这个题时,首先做了一个模拟,当n=3时,gray code应该是
000
001
011
010
110
100
101
111
看了半天,也没看出来什么规律。后来上网一查GrayCode(http://en.wikipedia.org/wiki/Gray_code)才发现原来推导的gray code顺序错了。第六个应该是111。
n=3时,正确的GrayCode应该是
000
001
011
010
110
111 //如果按照题意的话,只是要求有一位不同,这里也可以是100
101
100

这样的话,规律就出来了,n=k时的Gray Code,相当于n=k-1时的Gray Code的逆序 加上 1<<k。

[Code]
1:  vector<int> grayCode(int n) {  
2:      // Start typing your C/C++ solution below  
3:      // DO NOT write int main() function  
4:      vector<int> result;      
5:      result.push_back(0);  
6:      for(int i=0; i< n; i++)  
7:      {  
8:        int highestBit = 1<<i;  
9:        int len = result.size();  
10:        for(int i = len-1; i>=0; i--)  
11:        {  
12:          result.push_back(highestBit + result[i]);  
13:        }  
14:      }  
15:      return result;  
16:    }  

[总结]
题意不清楚,如果每次只是与上一个数有一个位不同的话,其实有很多种组合出来。如果不是查了Gray Code的定义,根本看不出来什么规律。
而且,Gray Code这种东西,必然有数学解,否则在早期的工程界是没法应用的。想了一下,其实也可以这么做,第i个数可以由如下公式产生: (i>>1)^i,所以代码也可以是:

1:  vector<int> grayCode(int n)   
2:    {   
3:      vector<int> ret;   
4:      int size = 1 << n;   
5:      for(int i = 0; i < size; ++i)   
6:        ret.push_back((i >> 1)^i);   
7:      return ret;   
8:    }   
不过这种数学解就失去了interview的意思了。




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