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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 [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.
这道题感觉是一个找规律的题目,找到规律后就很好求解,感觉不是一道回溯的题。
对于n=2,它的结果包括n=1时的结果左边补零,以及逆序遍历n=1时的结果左边补1,
规律如下图,列出了n=1,n=2,n=3时的情况。
根据这个规律如果已知n=k的情况,那么n=k+1的结果包括对n=k的结果左边补零,即保存不变,然后逆序遍历n=k的结果左边补1即可。
runtime:4ms
class Solution {
public:
vector<int> grayCode(int n) {
vector<int> result(1);
for(int i=0;i<n;i++)
{
for(int j=result.size()-1;j>=0;j--)
{
result.push_back((1<<i)+result[j]);
}
}
return result;
}
};
解法二就涉及到gray code的数学知识了,要是知道这个数学知识,可以在几分钟之内就解出这道题。
格雷码可以由对应的十进制数求出:grayCode=i^i>>1
runtime:4ms
class Solution {
public:
vector<int> grayCode(int n) {
vector<int> result;
for(int i=0;i<1<<n;i++)
{
result.push_back(i^i>>1);
}
return result;
}
};
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原文地址:http://blog.csdn.net/u012501459/article/details/46790683