[LeetCode] 89. Gray Code 格雷编码

 

An n-bit gray code sequence is a sequence of 2n integers where:

  • Every integer is in the inclusive range [0, 2n - 1],
  • The first integer is 0,
  • An integer appears no more than once in the sequence,
  • The binary representation of every pair of adjacent integers differs by exactly one bit, and
  • The binary representation of the first and last integers differs by exactly one bit.

Given an integer n, return any valid n-bit gray code sequence.

 

Example 1:

Input: n = 2
Output: [0,1,3,2]
Explanation:
The binary representation of [0,1,3,2] is [00,01,11,10].
- 00 and 01 differ by one bit
- 01 and 11 differ by one bit
- 11 and 10 differ by one bit
- 10 and 00 differ by one bit
[0,2,3,1] is also a valid gray code sequence, whose binary representation is [00,10,11,01].
- 00 and 10 differ by one bit
- 10 and 11 differ by one bit
- 11 and 01 differ by one bit
- 01 and 00 differ by one bit

Example 2:

Input: n = 1
Output: [0,1]

 

Constraints:

  • 1 <= n <= 16

 

这道题是关于格雷码的,猛地一看感觉完全没接触过格雷码,但是看了维基百科后,隐约的感觉原来好像哪门可提到过,哎全还给老师了。这道题如果不了解格雷码,还真不太好做,幸亏脑补了维基百科,上面说格雷码是一种循环二进制单位距离码,主要特点是两个相邻数的代码只有一位二进制数不同的编码,格雷码的处理主要是位操作 Bit Operation,LeetCode中关于位操作的题也挺常见,比如 Repeated DNA Sequences  Single Number, 和  Single Number II 等等。三位的格雷码和二进制数如下:

 

Int    Grey Code    Binary
 0      000        000
 1      001        001
 2      011        010
 3      010        011
 4      110        100
 5      111        101
 6      101        110
 7      100        111

 

其实这道题还有多种解法。首先来看一种最简单的,是用到格雷码和二进制数之间的相互转化,可参见我之前的博客 Convertion of grey code and binary 格雷码和二进制数之间的转换 ,明白了转换方法后,这道题完全没有难度,代码如下:

 

解法一:

// Binary to grey code
class Solution {
public:
    vector<int> grayCode(int n) {
        vector<int> res;
        for (int i = 0; i < pow(2,n); ++i) {
            res.push_back((i >> 1) ^ i);
        }
        return res;
    }
};

 

然后我们来看看其他的解法,参考维基百科上关于格雷码的性质,有一条是说镜面排列的,n位元的格雷码可以从 n-1 位元的格雷码以上下镜射后加上新位元的方式快速的得到,如下图所示一般。

有了这条性质,我们很容易的写出代码如下:

 

解法二:

// Mirror arrangement
class Solution {
public:
    vector<int> grayCode(int n) {
        vector<int> res{0};
        for (int i = 0; i < n; ++i) {
            int size = res.size();
            for (int j = size - 1; j >= 0; --j) {
                res.push_back(res[j] | (1 << i));
            }
        }
        return res;
    }
};

 

维基百科上还有一条格雷码的性质是直接排列,以二进制为0值的格雷码为第零项,第一项改变最右边的位元,第二项改变右起第一个为1的位元的左边位元,第三、四项方法同第一、二项,如此反复,即可排列出n个位元的格雷码。根据这条性质也可以写出代码,不过相比前面的略微复杂,代码如下:

0 0 0
0 0 1
0 1 1
0 1 0
1 1 0
1 1 1
1 0 1
1 0 0

 

解法三:

// Direct arrangement 
class Solution {
public:
    vector<int> grayCode(int n) {
        vector<int> res{0};
        int len = pow(2, n);
        for (int i = 1; i < len; ++i) {
            int pre = res.back();
            if (i % 2 == 1) {
                pre = (pre & (len - 2)) | ((~pre) & 1);
            } else {
                int cnt = 1, t = pre;
                while ((t & 1) != 1) {
                    ++cnt;
                    t >>= 1;
                }
                if ((pre & (1 << cnt)) == 0) pre |= (1 << cnt);
                else pre &= ~(1 << cnt);
            }
            res.push_back(pre);
        }
        return res;
    }
};

 

上面三种解法都需要事先了解格雷码及其性质,假如我们之前并没有接触过格雷码,那么其实也可以用比较笨的方法来找出结果,比如下面这种方法用到了一个 HashSet 来保存已经产生的结果,我们从0开始,遍历其二进制每一位,对其取反,然后看其是否在 HashSet 中出现过,如果没有,我们将其加入 HashSet 和结果 res 中,然后再对这个数的每一位进行遍历,以此类推就可以找出所有的格雷码了,参见代码如下:

 

解法四:

class Solution {
public:
    vector<int> grayCode(int n) {
        vector<int> res;
        unordered_set<int> s;
        helper(n, s, 0, res);
        return res;
    }
    void helper(int n, unordered_set<int>& s, int out, vector<int>& res) {
        if (!s.count(out)) {
            s.insert(out);
            res.push_back(out);
        }
        for (int i = 0; i < n; ++i) {
            int t = out;
            if ((t & (1 << i)) == 0) t |= (1 << i);
            else t &= ~(1 << i);
            if (s.count(t)) continue;
            helper(n, s, t, res);
            break;
        }
    }
};

 

既然递归方法可以实现,那么就有对应的迭代的写法,当然需要用 stack 来辅助,参见代码如下:

 

解法五:

class Solution {
public:
    vector<int> grayCode(int n) {
        vector<int> res{0};
        unordered_set<int> s;
        stack<int> st;
        st.push(0);
        s.insert(0);
        while (!st.empty()) {
            int t = st.top(); st.pop();
            for (int i = 0; i < n; ++i) {
                int k = t;
                if ((k & (1 << i)) == 0) k |= (1 << i);
                else k &= ~(1 << i);
                if (s.count(k)) continue;
                s.insert(k);
                st.push(k);
                res.push_back(k);
                break;
            }
        }
        return res;
    }
};

 

Github 同步地址:

https://github.com/grandyang/leetcode/issues/89

 

参考资料:

https://leetcode.com/problems/gray-code/

https://leetcode.com/problems/gray-code/discuss/29891/Share-my-solution

https://leetcode.com/problems/gray-code/discuss/29881/An-accepted-three-line-solution-in-JAVA

 

LeetCode All in One 题目讲解汇总(持续更新中...)

posted @ 2015-03-05 13:52  Grandyang  阅读(20079)  评论(3编辑  收藏  举报
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