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117 lines
2.6 KiB
Markdown
Executable File
117 lines
2.6 KiB
Markdown
Executable File
# [89. Gray Code](https://leetcode.com/problems/gray-code/)
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## 题目
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The gray code is a binary numeral system where two successive values differ in only one bit.
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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.
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**Example 1**:
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Input: 2
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Output: [0,1,3,2]
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Explanation:
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00 - 0
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01 - 1
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11 - 3
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10 - 2
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For a given n, a gray code sequence may not be uniquely defined.
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For example, [0,2,3,1] is also a valid gray code sequence.
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00 - 0
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10 - 2
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11 - 3
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01 - 1
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**Example 2**:
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Input: 0
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Output: [0]
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Explanation: We define the gray code sequence to begin with 0.
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A gray code sequence of n has size = 2n, which for n = 0 the size is 20 = 1.
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Therefore, for n = 0 the gray code sequence is [0].
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## 题目大意
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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。
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## 解题思路
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- 输出 n 位格雷码
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- 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。
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- 可以直接模拟，也可以用递归求解。
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## 代码
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```go
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package leetcode
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// 解法一 递归方法，时间复杂度和空间复杂度都较优
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func grayCode(n int) []int {
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if n == 0 {
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return []int{0}
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}
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res := []int{}
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num := make([]int, n)
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generateGrayCode(int(1<<uint(n)), 0, &num, &res)
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return res
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}
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func generateGrayCode(n, step int, num *[]int, res *[]int) {
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if n == 0 {
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return
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}
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*res = append(*res, convertBinary(*num))
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if step%2 == 0 {
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(*num)[len(*num)-1] = flipGrayCode((*num)[len(*num)-1])
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} else {
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index := len(*num) - 1
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for ; index >= 0; index-- {
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if (*num)[index] == 1 {
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break
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}
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}
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if index == 0 {
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(*num)[len(*num)-1] = flipGrayCode((*num)[len(*num)-1])
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} else {
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(*num)[index-1] = flipGrayCode((*num)[index-1])
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}
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}
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generateGrayCode(n-1, step+1, num, res)
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return
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}
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func convertBinary(num []int) int {
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res, rad := 0, 1
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for i := len(num) - 1; i >= 0; i-- {
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res += num[i] * rad
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rad *= 2
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}
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return res
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}
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func flipGrayCode(num int) int {
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if num == 0 {
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return 1
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}
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return 0
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}
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// 解法二 直译
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func grayCode1(n int) []int {
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var l uint = 1 << uint(n)
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out := make([]int, l)
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for i := uint(0); i < l; i++ {
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out[i] = int((i >> 1) ^ i)
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}
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return out
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}
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``` |