LeetCode-Gray Code
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.
class Solution {
public:
inline int changeBit(int value,int i){
unsigned int b=1;
i--;
b=b<<i;
if((b&value)>0){
return value&(~b);
}
else{
return value|b;
}
return value;
}
inline int differ(int a,int b){
a=a^b;
int i=1;
for(;;){
if(a&1){
return i;
}
else{
i++;
a=a>>1;
}
}
}
vector<int> grayCode(int n) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
vector<int>ret;
if(n<0){
return ret;
}
else if(n==0){
ret.push_back(0);
return ret;
}
int length=1;
int b=n;
for(;b>0;b--)length*=2;
int* bits=new int[length];
memset(bits,0,length*sizeof(int)/sizeof(char));
ret.push_back(0);
ret.push_back(1);
int current=1;
if(n==1){
delete bits;
return ret;
}
bits[0]=1;
bits[current]++;
bool forward=true;
int i=1;
for(;;){
/*for(int j=0;j<ret.size();j++){
cout<<ret[j]<<" ";
}
cout<<endl;*/
if(forward){
int v=changeBit(ret[current],i);
current++;
ret.push_back(v);
if(bits[ret[current]]){
forward=false;
}
else{
forward=true;
i=1;
if(current==length-1)
break;
}
bits[ret[current]]++;
}
else{
bits[ret[current]]--;
current--;
i=differ(ret[current],ret[current+1]);
ret.pop_back();
if(i==n){
forward=false;
}
else {
forward=true;
i++;
}
}
}
delete bits;
return ret;
}
};
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