[LeetCode] 89. Gray Code 格雷码

2021年09月15日 阅读数:1
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The gray code is a binary numeral system where two successive values differ in only one bit.html

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.java

Example 1:python

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:app

Input: 0
Output: [0]
Explanation: We define the gray code sequence to begin with 0.
             A gray sequence of n has size = 2n, which for n = 0 the size is 20 = 1.
             Therefore, for n = 0 the gray code sequence is [0].

首先要清楚什么是格雷码,而后用位操做来处理。post

Java:url

public List<Integer> grayCode(int n) {
    List<Integer> result = new LinkedList<>();
    for (int i = 0; i < 1<<n; i++) result.add(i ^ i>>1);
    return result;
}

Java:code

public List<Integer> grayCode(int n) {
    List<Integer> rs=new ArrayList<Integer>();
    rs.add(0);
    for(int i=0;i<n;i++){
        int size=rs.size();
        for(int k=size-1;k>=0;k--)
            rs.add(rs.get(k) | 1<<i);
    }
    return rs;
}    

Python:orm

class Solution(object):
    def grayCode(self, n):
        """
        :type n: int
        :rtype: List[int]
        """
        result = [0]
        for i in xrange(n):
            for n in reversed(result):
                result.append(1 << i | n)
        return result

Python:htm

# Proof of closed form formula could be found here:
# http://math.stackexchange.com/questions/425894/proof-of-closed-form-formula-to-convert-a-binary-number-to-its-gray-code
class Solution2(object):
    def grayCode(self, n):
        """
        :type n: int
        :rtype: List[int]
        """
        return [i >> 1 ^ i for i in xrange(1 << n)]

C++:blog

// Time:  (2^n)
// Space: O(1)
class Solution {
public:
    vector<int> grayCode(int n) {
        vector<int> result = {0};
        for (int i = 0; i < n; ++i) {
            for (int j = result.size() - 1; j >= 0; --j) {
                result.emplace_back(1 << i | result[j]);
            }
        }
        return result;
    }
};

C++:  

// Time:  (2^n)
// Space: O(1)
// Proof of closed form formula could be found here:
// http://math.stackexchange.com/questions/425894/proof-of-closed-form-formula-to-convert-a-binary-number-to-its-gray-code
class Solution2 {
public:
    vector<int> grayCode(int n) {
        vector<int> result;
        for (int i = 0; i < 1 << n; ++i) {
            result.emplace_back(i >> 1 ^ i);
        }
        return result;
    }
};

  

    

 

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