[LintCode] Gray Code

cocopeak 发布于2019-08-14 15:15 / 2846人阅读

摘要：参考了的算法，简化了一下每个循环更新对应最高位是第位，就增加个数为倒序循环次，会镜像提取上一次循环产生的结果镜像镜像镜像

Problem

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, find the sequence of gray code. A gray code sequence must begin with 0 and with cover all 2n integers.
http://mathworld.wolfram.com/...

Example

Given n = 2, return [0,1,3,2]. Its gray code sequence is:

Solution

其实就是找规律，0-01-0132－01326754这样，每个循环(i)：

增加当前res.size()个新数；

新的循环先在2进制的第(i+1)位加1，同时与之前res所存的数字倒序相加产生新数；

存入新数，进入下一个循环后更新size。

参考了codesolutiony的算法，简化了一下：

public class Solution {
    public ArrayList grayCode(int n) {
        ArrayList res = new ArrayList();
        res.add(0);
        for (int i = 0; i < n; i++) {
            //每个循环更新size: 1, 3, 7...
            int size = res.size() - 1;
            //i对应最高位是第i+1位，res就增加2^(i+1)个数：(1<= 0; j--) {
                res.add((1<
Recursion
public class Solution {
    public List grayCode(int n) {
        List res = new ArrayList();
        if (n == 0) {
            res.add(0);
            return res;
        }
        res = grayCode(n-1);
        for (int i = res.size()-1; i >= 0; i--) {
            int num = res.get(i);
            res.add(num+(1<<(n-1)));
        }
        return res;
    }
}
Using XOR
public class Solution {
    public List grayCode(int n) {
        List res = new ArrayList();
        for(int i = 0; i < Math.pow(2,n); i++) res.add(i >> 1 ^ i);
        return res;
    }
}

GPU云服务器云服务器 Gray LintCode LintCode134_LRU Code

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