leetcode 89. Gray Code
2016-04-17 20:59
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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
is:
Note:
For a given n, a gray code sequence is not uniquely defined.
For example,
For now, the judge is able to judge based on one instance of gray code sequence. Sorry about that.
这种方法基于格雷码是反射码的事实,利用递归的如下规则来构造:
1位格雷码有两个码字
(n+1)位格雷码中的前2n个码字等于n位格雷码的码字,按顺序书写,加前缀0
(n+1)位格雷码中的后2n个码字等于n位格雷码的码字,按逆序书写,加前缀1[3]
n+1位格雷码的集合 = n位格雷码集合(顺序)加前缀0 + n位格雷码集合(逆序)加前缀1
class Solution {
vector<vector<int>> getGray(int n) {
vector<vector<int>>re;
if(n==0)
{
vector<int>aa;
aa.push_back(0);
re.push_back(aa);
return re;
}
if (n == 1)
{
vector<int>aa, bb;
aa.push_back(0);
bb.push_back(1);
re.push_back(aa); re.push_back(bb);
return re;
}
vector<vector<int>>rr = getGray(n - 1);
for (int i = 0; i < rr.size(); i++)
{
vector<int>ss = rr[i];
ss.insert(ss.begin(), 0);
re.push_back(ss);
}
for (int i = rr.size() - 1; i >= 0; i--)
{
vector<int>ss = rr[i];
ss.insert(ss.begin(), 1);
re.push_back(ss);
1 << 2;
}
return re;
}
public:
vector<int> grayCode(int n) {
vector<vector<int>>rr = getGray(n);
vector<int>re;
for (int i = 0; i < rr.size(); i++)
{
int dd = 0;
for (int j = 0; j < n; j++)
dd += (rr[i][j] << (n - 1 - j));
re.push_back(dd);
}
return re;
}
};
accept
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.
递归生成码表
这种方法基于格雷码是反射码的事实,利用递归的如下规则来构造:1位格雷码有两个码字
(n+1)位格雷码中的前2n个码字等于n位格雷码的码字,按顺序书写,加前缀0
(n+1)位格雷码中的后2n个码字等于n位格雷码的码字,按逆序书写,加前缀1[3]
n+1位格雷码的集合 = n位格雷码集合(顺序)加前缀0 + n位格雷码集合(逆序)加前缀1
2位格雷码 | 3位格雷码 | 4位格雷码 | 4位自然二进制码 |
---|---|---|---|
00 01 11 10 | 000 001 011 010 110 111 101 100 | 0000 0001 0011 0010 0110 0111 0101 0100 1100 1101 1111 1110 1010 1011 1001 1000 | 0000 0001 0010 0011 0100 0101 0110 0111 1000 1001 1010 1011 1100 1101 1110 1111 |
vector<vector<int>> getGray(int n) {
vector<vector<int>>re;
if(n==0)
{
vector<int>aa;
aa.push_back(0);
re.push_back(aa);
return re;
}
if (n == 1)
{
vector<int>aa, bb;
aa.push_back(0);
bb.push_back(1);
re.push_back(aa); re.push_back(bb);
return re;
}
vector<vector<int>>rr = getGray(n - 1);
for (int i = 0; i < rr.size(); i++)
{
vector<int>ss = rr[i];
ss.insert(ss.begin(), 0);
re.push_back(ss);
}
for (int i = rr.size() - 1; i >= 0; i--)
{
vector<int>ss = rr[i];
ss.insert(ss.begin(), 1);
re.push_back(ss);
1 << 2;
}
return re;
}
public:
vector<int> grayCode(int n) {
vector<vector<int>>rr = getGray(n);
vector<int>re;
for (int i = 0; i < rr.size(); i++)
{
int dd = 0;
for (int j = 0; j < n; j++)
dd += (rr[i][j] << (n - 1 - j));
re.push_back(dd);
}
return re;
}
};
accept
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