0089. Gray Code

89. Gray Code #

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.

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].

题目大意 #

格雷编码是一个二进制数字系统,在该系统中,两个连续的数值仅有一个位数的差异。给定一个代表编码总位数的非负整数 n,打印其格雷编码序列。格雷编码序列必须以 0 开头。

解题思路 #

  • 输出 n 位格雷码
  • 格雷码生成规则:以二进制为0值的格雷码为第零项,第一次改变最右边的位元,第二次改变右起第一个为1的位元的左边位元,第三、四次方法同第一、二次,如此反复,即可排列出 n 个位元的格雷码。
  • 可以直接模拟,也可以用递归求解。

代码 #


package leetcode

// 解法一 递归方法,时间复杂度和空间复杂度都较优
func grayCode(n int) []int {
	if n == 0 {
		return []int{0}
	}
	res := []int{}
	num := make([]int, n)
	generateGrayCode(int(1<<uint(n)), 0, &num, &res)
	return res
}

func generateGrayCode(n, step int, num *[]int, res *[]int) {
	if n == 0 {
		return
	}
	*res = append(*res, convertBinary(*num))

	if step%2 == 0 {
		(*num)[len(*num)-1] = flipGrayCode((*num)[len(*num)-1])
	} else {
		index := len(*num) - 1
		for ; index >= 0; index-- {
			if (*num)[index] == 1 {
				break
			}
		}
		if index == 0 {
			(*num)[len(*num)-1] = flipGrayCode((*num)[len(*num)-1])
		} else {
			(*num)[index-1] = flipGrayCode((*num)[index-1])
		}
	}
	generateGrayCode(n-1, step+1, num, res)
	return
}

func convertBinary(num []int) int {
	res, rad := 0, 1
	for i := len(num) - 1; i >= 0; i-- {
		res += num[i] * rad
		rad *= 2
	}
	return res
}

func flipGrayCode(num int) int {
	if num == 0 {
		return 1
	}
	return 0
}

// 解法二 直译
func grayCode1(n int) []int {
	var l uint = 1 << uint(n)
	out := make([]int, l)
	for i := uint(0); i < l; i++ {
		out[i] = int((i >> 1) ^ i)
	}
	return out
}


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