• Gray Code


    题目:

    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.

    就是把n位二进制数都转成十进制返回。代码:

     1     vector<int> grayCode(int n) {
     2         vector<int> result;
     3         if(!n) {
     4             result.push_back(0);
     5             return result;
     6         }
     7         bool* bits = (bool*)malloc(n);
     8         memset(bits,0,n);
     9         getGrayCode(result,n,0,bits);
    10         return result;
    11     }
    12     void getGrayCode(vector<int>& result,int n,int i,bool* bits){
    13         if(n==i){
    14             result.push_back(b2i(bits,n));
    15             return ;
    16         }
    17         getGrayCode(result,n,i+1,bits);
    18         bits[i]=!bits[i];
    19         getGrayCode(result,n,i+1,bits);
    20     }
    21     int b2i(bool* bits,int n){
    22         int result=0,c=0;
    23         for(int i=n-1;i>=0;i--){
    24             if(!c) c=1;
    25             else c*=2;
    26             if(bits[i]){
    27                 result += c;
    28             }
    29         }
    30         return result;
    31     }

    时间复杂度太大了,以为过不了,结果还是过了,都2n*n了。不过规模已经是指数级,没办法降啊。
    后来看到这个解法,于是去wikipedia上面看了下,确实非常简单,下面的链接:
    http://blog.csdn.net/doc_sgl/article/details/12251523

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  • 原文地址:https://www.cnblogs.com/mike442144/p/3475451.html
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