Leetcode 89. 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.

s思路：

1. 方法１：iterative。双重循环，外循环负责增加宽度，内循环复责增加长度。

2. 方法２：recursive，做的过程是:为得到n比特的gray code,先构造n-1比特，所以整个过程是可迭代的，且迭代最底层是n==0,return {0};

3. 做的过程是先想到iterative,反而忽略了recursive。其实越是规则的构造，用recursive越方便简单！复杂的构造，还是iterative好用！不同的题，不同的方法，不过最好都试一试，建立一种直观的sense。每次遇到题，依靠sense就可以解决！

//方法１：iterative。双重循环。
class Solution {
public:
vector<int> grayCode(int n) {
//
vector<int> res;
if(n<0) return res;
res.push_back(0);
int j=0;
while(j<n){
int i=res.size()-1;
while(i>=0){
res.push_back(res[i]+(1<<j));
i--;
}
j++;
}
return res;
}
};

//方法２：用recursive的方法来做！参考了以前做的
class Solution {
public:
vector<int> grayCode(int n) {
//
if(n==0) return {0};
vector<int> res=grayCode(n-1);
int n0=res.size();
res.resize(1<<n);
int n1=res.size();
for(int i=n0;i<n1;i++){
res[i]=res[--n0]+(1<<(n-1));
}
return res;
}
};