LeetCode #89 - Gray Code

最新推荐文章于 2021-08-27 21:40:34 发布

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本文介绍了一种生成格雷码序列的有效算法。通过递推方法，从n-1位到n位，按特定规则扩展序列，确保每两个连续值只有一位不同。文章提供了C++实现代码，并给出具体示例，帮助理解格雷码及其生成过程。

题目描述：

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.

Example 1:

Input: 2
Output: [0,1,3,2]
Explanation:
00 - 0
01 - 1
11 - 3
10 - 2

For a given n, a gray code sequence may not be uniquely defined.
For example, [0,2,3,1] is also a valid gray code sequence.

00 - 0
10 - 2
11 - 3
01 - 1

Example 2:

Input: 0
Output: [0]
Explanation: We define the gray code sequence to begin with 0.
             A gray code sequence of n has size = 2^n, which for n = 0 the size is 2^0 = 1.
             Therefore, for n = 0 the gray code sequence is [0].

给定n，求2^n个格雷码序列。格雷码从n-1位到n位递推的时候，就是按照以前的序列将第n个高位置为1，即增加2^n，得到新的序列，所以是一个循环递推的过程。

class Solution {
public:
    vector<int> grayCode(int n) {
        vector<int> result;
        result.push_back(0);
        for(int i=0;i<n;i++)
        {
            vector<int> temp=result;
            for(int j=temp.size()-1;j>=0;j--)
                result.push_back(temp[j]+pow(2,i));
        }
        return result;
    }
};