89. Gray Code (Medium)

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

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 = 2n, which for n = 0 the size is 20 = 1.
             Therefore, for n = 0 the gray code sequence is [0].

Solutions

class Solution {
    public List<Integer> grayCode(int n) {
        List<Integer> ans = new ArrayList<>();
        for (int i = 0; i < (1 << n); i++) {
            ans.add(i ^ (i >> 1));
        }

        // assume x is even, x=?0, y=x+1, then y=?1, odd, so that x^y=0[0,k]1
        // assume x is odd, x=?01[1,j], y=x+1, then y=?10[1,j], even, so that x^y=0[0,k]11[1,j]

        // assume i-1 is is even, such that i is odd and (i-1)>>1 = i>>1, i, i-1 oly differ in the last bit.
        // then ans.get(i)^ans.get(i-1)=i^(i>>1)^(i-1)^((i-1)>>1)=i^(i-1)^(i>>1)^((i-1)>>1)=1^0=1
        // we can see the only difference is the last bit

        // assume i-1 is is odd, such that i is even and (i-1)>>1 = (i>>1)-1
        // then ans.get(i)^ans.get(i-1)=i^(i>>1)^(i-1)^((i-1)>>1)=i^(i-1)^(i>>1)^((i-1)>>1)=11[1,k]^01[1,k]=10[1,k]
        // we can see the only difference is the first bit

        return ans;
    }
}

Incorrect Solutions

References

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