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[LintCode] Gray Code

cocopeak / 2847人閱讀

摘要：參考了的算法，簡(jiǎn)化了一下每個(gè)循環(huán)更新對(duì)應(yīng)最高位是第位，就增加個(gè)數(shù)為倒序循環(huán)次，會(huì)鏡像提取上一次循環(huán)產(chǎn)生的結(jié)果鏡像鏡像鏡像

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:

00 - 0
01 - 1
11 - 3
10 - 2
Solution

其實(shí)就是找規(guī)律，0-01-0132－01326754這樣，每個(gè)循環(huán)(i)：

增加當(dāng)前res.size()個(gè)新數(shù)；

新的循環(huán)先在2進(jìn)制的第(i+1)位加1，同時(shí)與之前res所存的數(shù)字倒序相加產(chǎn)生新數(shù)；

存入新數(shù)，進(jìn)入下一個(gè)循環(huán)后更新size。

參考了codesolutiony的算法，簡(jiǎn)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++) {
            //每個(gè)循環(huán)更新size: 1, 3, 7...
            int size = res.size() - 1;
            //i對(duì)應(yīng)最高位是第i+1位，res就增加2^(i+1)個(gè)數(shù)：(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;
    }
}

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