题目描述
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.
For example, given n = 2, return[0,1,3,2]. Its gray code sequence is:
00 - 0
01 - 1
11 - 3
10 - 2
Note:
For a given n, a gray code sequence is not uniquely defined.
For example,[0,2,3,1]is also a valid gray code sequence according to the above definition.
For now, the judge is able to judge based on one instance of gray code sequence. Sorry about that.
- 思路
求 n 位二进制的所有组合情况，且任意两个相邻的串只有一位不相同。
即格雷码，两种方法。
一是公式：(i >> 1) ^ i，其中 i 为位数
二是模拟实现，从2位开始每位是前一位的（格雷码）变化，加0的情况就是上一位的值集（大小和顺序均不变），加1就是变为二进制后首位加1，其他位不变，注意是逆序添加。
//方法一
import java.util.ArrayList;
public class Solution {
public ArrayList<Integer> grayCode(int n) {
ArrayList<Integer> list = new ArrayList<Integer>();
if(n < 0)
return list;
n = (int)Math.pow(2, n);
for(int i=0; i<n; i++){
list.add((i >> 1) ^ i);
}
return list;
}
}
//方法二
import java.util.ArrayList;
public class Solution {
public ArrayList<Integer> grayCode(int n) {
ArrayList<Integer> list = new ArrayList<Integer>();
if(n < 0)
return list;
if(n >= 0)
list.add(0);
if(n >= 1)
list.add(1);
for(int i=2; i<=n; i++){ //位数
int size = list.size();
for(int j=size-1; j>=0; j--){ //逆序给每个数的二进制前面加1（加0的情况与不加一样，省略）
list.add(list.get(j) + (1 << i-1));
}
}
return list;
}
}
