### 题目

There are 8 prison cells in a row, and each cell is either occupied or vacant.

Each day, whether the cell is occupied or vacant changes according to the following rules:

If a cell has two adjacent neighbors that are both occupied or both vacant, then the cell becomes occupied. Otherwise, it becomes vacant. (Note that because the prison is a row, the first and the last cells in the row can’t have two adjacent neighbors.)

We describe the current state of the prison in the following way: cells[i] == 1 if the i-th cell is occupied, else cells[i] == 0.

Given the initial state of the prison, return the state of the prison after N days (and N such changes described above.)

Example 1:

Input: cells = [0,1,0,1,1,0,0,1], N = 7
Output: [0,0,1,1,0,0,0,0]
Explanation:
The following table summarizes the state of the prison on each day:
Day 0: [0, 1, 0, 1, 1, 0, 0, 1]
Day 1: [0, 1, 1, 0, 0, 0, 0, 0]
Day 2: [0, 0, 0, 0, 1, 1, 1, 0]
Day 3: [0, 1, 1, 0, 0, 1, 0, 0]
Day 4: [0, 0, 0, 0, 0, 1, 0, 0]
Day 5: [0, 1, 1, 1, 0, 1, 0, 0]
Day 6: [0, 0, 1, 0, 1, 1, 0, 0]
Day 7: [0, 0, 1, 1, 0, 0, 0, 0]


Example 2:

Input: cells = [1,0,0,1,0,0,1,0], N = 1000000000
Output: [0,0,1,1,1,1,1,0]


Note:

• cells.length == 8
• cells[i] is in {0, 1}
• 1 <= N <= 10^9

### 最朴素的做法，直接在数组上操作

today:      [0, 1, 0, 1, 1, 0, 0, 1]
\  X  X  X  X  X  /
tomorrow:   [0, 1, 1, 0, 0, 0, 0, 0]
|
today[0]和today[2]决定了tomorrow[1]


#### 代码

class Solution {
public int[] prisonAfterNDays(int[] cells, int N) {
for (int i = 1; i <= N; i++) {
int[] next = change(cells);
cells = next;
}
return cells;
}

private int[] change(int[] pre) {
int[] curr = new int[pre.length];
for (int i = 1; i < pre.length - 1; i++) {
curr[i] = (pre[i - 1] == pre[i + 1])? 1 : 0;
}
return curr;
}
}


### 位操作

0011
1010 ^
------
1001 ~
------
0110


#### 代码

class Solution {
public int[] prisonAfterNDays(int[] cells, int N) {
int num = 0;
for (int i = 0; i < cells.length; i++) {
num <<= 1;
num |= cells[i];
}
for (int i = 1; i <= N; i++) {
num = change(num);
}
int[] res = new int[cells.length];
for (int i = cells.length - 1; i >= 0; i--) {
res[i] = num & 1;
num >>= 1;
}
return res;
}

private int change(int pre) {
int mask = 126; // 126 = 0111 1110
int left = pre << 1;
int right = pre >> 1;
int res = (~ (left ^ right)) & mask;
return res;
}
}


### 结果是循环的

Day 0:  [0, 1, 0, 1, 1, 0, 0, 1]
--------------------------------
Day 1:  [0, 1, 1, 0, 0, 0, 0, 0] --+
Day 2:  [0, 0, 0, 0, 1, 1, 1, 0]   |
Day 3:  [0, 1, 1, 0, 0, 1, 0, 0]   |
Day 4:  [0, 0, 0, 0, 0, 1, 0, 0]   |
Day 5:  [0, 1, 1, 1, 0, 1, 0, 0]   |
Day 6:  [0, 0, 1, 0, 1, 1, 0, 0]   |
Day 7:  [0, 0, 1, 1, 0, 0, 0, 0]   |
Day 8:  [0, 0, 0, 0, 0, 1, 1, 0]   | 循
Day 9:  [0, 1, 1, 1, 0, 0, 0, 0]   | 环
Day 10: [0, 0, 1, 0, 0, 1, 1, 0]   |
Day 11: [0, 0, 1, 0, 0, 0, 0, 0]   |
Day 12: [0, 0, 1, 0, 1, 1, 1, 0]   |
Day 13: [0, 0, 1, 1, 0, 1, 0, 0]   |
Day 14: [0, 0, 0, 0, 1, 1, 0, 0]   |
--------------------------------   |
Day 15: [0, 1, 1, 0, 0, 0, 0, 0] <-+
...
...


#### 代码

class Solution {
public int[] prisonAfterNDays(int[] cells, int N) {
int num = 0;
for (int i = 0; i < cells.length; i++) {
num <<= 1;
num |= cells[i];
}
num = change(num);
int times = (N - 1) % 14;
for (int i = 0; i < times; i++) {
num = change(num);
}
int[] res = new int[cells.length];
for (int i = cells.length - 1; i >= 0; i--) {
res[i] = num & 1;
num >>= 1;
}
return res;
}

private int change(int pre) {
int mask = 126; // 126 = 0111 1110
int left = pre << 1;
int right = pre >> 1;
int res = (~ (left ^ right)) & mask;
return res;
}
}