M
tags: Backtracking
TODO:
1. backtracking, using set to perform contains()
2. BFS: use queue to keep the mutations
题目蛋疼，目前只接受一种结果。
BackTracking + DFS:
Recursive helper里每次flip一个 自己/左边/右边. Flip过后还要恢复原样.遍历所有.
曾用法（未仔细验证）：
基本想法就是从一个点开始往一个方向走，每次flip一个bit, 碰壁的时候就回头走。
```
/*
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.
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*/
//Another solution, 02.10.2016 using DFS
//generate order, output the numerical value of each binary code. Integer.parseInt("10101", 2).
//Start with n-bit char[] of 0's. Flip one bit at a time.
//Recursive helper. char[], index. Flip or not flip. DFS
public class Solution {
public List grayCode(int n) {
List rst = new ArrayList();
if (n < 0) {
return rst;
} else if (n == 0) {
rst.add(0);
return rst;
}
char[] list = new char[n];
for (int i = 0; i < n; i++) {
list[i] = '0';
}
helper(rst, list, n - 1);
return rst;
}
public void helper(List rst, char[] list, int index) {
rst.add(Integer.parseInt(new String(list), 2));
//self
list[index] = list[index] == '0' ? '1' : '0';
int num = Integer.parseInt(new String(list), 2);
if (!rst.contains(num)) {
helper(rst, list, index);
}
list[index] = list[index] == '0' ? '1' : '0';
//left
if (index -1 >= 0) {
list[index - 1] = list[index - 1] == '0' ? '1' : '0';
num = Integer.parseInt(new String(list), 2);
if (!rst.contains(num)) {
helper(rst, list, index - 1);
}
list[index - 1] = list[index - 1] == '0' ? '1' : '0';
}
//right
if (index + 1 < list.length) {
list[index + 1] = list[index + 1] == '0' ? '1' : '0';
num = Integer.parseInt(new String(list), 2);
if (!rst.contains(num)) {
helper(rst, list, index + 1);
}
list[index + 1] = list[index + 1] == '0' ? '1' : '0';
}
}
}
/*
Leetcode tags shows backtracking. That should be different approach than what I hvae below:
*/
/*
TRY: My code works for this run-through, however does not fit the OJ yet
0 0 0 [start, noting happend, flip index 0]
0 0 <-1 [move to flip left adjacent]
0 <-1 1 [move to flip left adjacent]
1-> 1 1 [move to flip right adjacent]
1 0-> 1 [move to flip right adjacent]
1 0 <-0 [move to flip left adjacent]
1 <-1 0 [move to flip left adjacent]
0 1 0 [done]
Conclusion: hit the wall and flip the other direction.
Every flip, add integer to list
Convert the char[] to string, then Integer.parse(str, 2) to integer
Simulate the steps:
For example, when n = 3, step = n - 1. It takes 2 steps from right side to reach left side, it hits the wall and turn around.
Now:
1. Initialize char[] and add '000'
2. do for loop on 1 ~ 2^n -2. last step '010' is stepped into, but no further action, so take 2^3 - 1 = 7 steps.
*/
public class Solution {
public List grayCode(int n) {
List rst = new ArrayList();
if (n < 0) {
return rst;
}
char[] bits = new char[n];
for (int i = 0; i < bits.length; i++) {
bits[i] = '0';
}
String str = new String(bits);
if (n == 0) {
str = "0";
}
rst.add(Integer.parseInt(str, 2));
int step = n - 1;
boolean LR = true;//L: true; R: false
int steps = (int)Math.pow(2, n) - 1;
for (int i = 0; i < steps; i++) {
bits[step] = bits[step] == '0' ? '1' : '0';
str = new String(bits);
rst.add(Integer.parseInt(str, 2));
if (LR) {
step--;
} else {
step++;
}
if (step == (n - 1) || step == 0) {//Turn around
LR = !LR;
}
}
return rst;
}
}
```