LeetCode 题解（158）: Gray Code

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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.

题解：

当前Gray Code等于前一个Gray Code的逆序加1 << n - 1。

C++版：

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

Java版：

public class Solution {public List<Integer> grayCode(int n) {List<Integer> result = new ArrayList<Integer>();result.add(0);if(n == 0)return result;result.add(1);for(int i = 1; i < n; i++) {for(int j = result.size() - 1; j >= 0; j--) {result.add(result.get(j) + (1 << i));}}return result;}
}

Python版：

class Solution:# @param {integer} n# @return {integer[]}def grayCode(self, n):if n == 0:return [0]result = [0, 1]for i in range(2, n+1):for j in range(len(result) - 1, -1, -1):result.append((1 << i - 1) + result[j])return result

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