Leetcode: Gray Code
2014-10-04 13:10
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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
Its gray code sequence is:
Note:
For a given n, a gray code sequence is not uniquely defined.
For example,
according to the above definition.
Assume the result always start with [0, 1]. For n = 2, to get the next gray number, add 10 to 01, that is 11, and again add 10 to 00, get the next number 10. Because adding the same value to adjacent numbers still gets adjacent numbers. Result for n
= 3 starts with results for n = 2, so on the basis of n = 2, add 100 to each number from bottom up, get the result. The complexity is O(2^n) time and O(2^n) space.
public class Solution {
public ArrayList<Integer> grayCode(int n) {
ArrayList<Integer> res = new ArrayList<Integer>();
if (n < 0) {
return res;
}
if (n == 0) {
res.add(0);
return res;
}
res.add(0);
res.add(1);
for (int i = 2; i <= n; i++) {
int size = res.size();
for (int j = size - 1; j >= 0; j--) {
res.add(res.get(j) + (1 << (i - 1)));
}
}
return res;
}
}
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.
Assume the result always start with [0, 1]. For n = 2, to get the next gray number, add 10 to 01, that is 11, and again add 10 to 00, get the next number 10. Because adding the same value to adjacent numbers still gets adjacent numbers. Result for n
= 3 starts with results for n = 2, so on the basis of n = 2, add 100 to each number from bottom up, get the result. The complexity is O(2^n) time and O(2^n) space.
public class Solution {
public ArrayList<Integer> grayCode(int n) {
ArrayList<Integer> res = new ArrayList<Integer>();
if (n < 0) {
return res;
}
if (n == 0) {
res.add(0);
return res;
}
res.add(0);
res.add(1);
for (int i = 2; i <= n; i++) {
int size = res.size();
for (int j = size - 1; j >= 0; j--) {
res.add(res.get(j) + (1 << (i - 1)));
}
}
return res;
}
}
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