[Leetcode] 89. Gray Code 解题报告

题目：

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.

思路：

总结该题目的规律：假设对于(n-1)而言，我们已经计算出来了gray code的序列grayCode(n-1)，那么对于n而言，一个可行的办法是什么呢？由于grayCode(n)中的每个元素都有n个二进制位，所以我们可以将grayCode(n)分为两部分，第一部分是首位为0的元素，第二部分是首位位1的元素。可以发现，其中第一部分恰好就是grayCode(n-1)中的元素。那么如果我们依次将第一部分的元素首位都变成1，然后逆序加入序列，不就构成了grayCode(n)的结果吗？下面的代码正是这一思路的结果，我们用递归实现了这一算法。

代码：

class Solution {
public:
    vector<int> grayCode(int n) {
        vector<int> result = {0};
        if (n == 0) {
            return result;
        }
        else if (n == 1)
        {
            result.push_back(1);
            return result;
        }
        result = grayCode(n-1);
        int half_size = result.size();
        int added_value = 1 << (n - 1);
        for(int i = half_size - 1; i >= 0; --i) {
            result.push_back(added_value + result[i]);
        }
        return result;
    }
};