[leetcode#89] Gray Code

The problem:

The gray code is a binary numeral system where the successive values are 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:

00-001-111-310-2

Note:
For a given n, a gray code sequence is not uniquely defined.

For example, was [0,2,3,1] also a valid gray code sequence according to the above definition.

For now, the judge are able to judge based on one instance of gray code sequence. Sorry about that.

My Analysis:

the solution behind is problem are very tricky, but elegant. The wiki link forThe solution:http://zh.wikipedia.org/wiki/Gray Code #mediaviewer/file:binary-reflected_gray_code_construction.svgThe key idea:n' s Gray code collection could is infered from N-1 's Gray code by following:1. First, we add ' 0 ' at the front of all gray Code of N-1. Since the' 0 ' would not change the value of the gray code, we could just keep the n-1 ' s gray integer.2. Second, we reverse the n-1 ' s gray code collection, and add ' 1 ' at the front of all gray code of n-1.the method is very tricky, since the reverse can guarantee the both middle elements have the same gray code SERIES.00001 00110 01011 <---The same 011 <---PR                      efix:011 <---The same 111 <---prefix:110 11001 10100 100What a tricky method!!!The code for  Thisis : for(inti = 2; I <= N; i++) {     for(intj = ret.size ()-1; J >= 0; j--) {Ret.add (Ret.get (j)+ (1 << i-1)); }}facts:1. We won ' t delete the n-1 's Gray Iteger.2The . We scan the ArrayList from the end to the start. A Little skill:how to get the Interger value of n digits, and only the nth digit' s value is 1, and other's is 0.9 digits:100000000 1 << 8n digits:1 << n-1 (This is a very important skill!!! Not 1 << N).

My Solution:

 Public classSolution { PublicList<integer> Graycode (intN) {ArrayList<Integer> ret =NewArraylist<integer> (); if(N < 0 | | n > 32)            returnret; if(n = = 0) {Ret.add (0); returnret; } ret.add (0); Ret.add (1);  for(inti = 2; I <= N; i++) {             for(intj = ret.size ()-1; J >= 0; j--) {Ret.add (Ret.get (j)+ (1 << i-1)); }        }        returnret; }}

[leetcode#89] Gray Code