LC 957. Prison Cells After N Days

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

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]

题目要求：8个（0，1）一排，两边相同中间变1，两边不同中间变0。问N次后的数组样子。

思路1：使用字典记录每一个过程和遍历时的N，如果有重复直接取模。减少运算量。

  class Solution {
  public:
  vector<int> prisonAfterNDays(vector<int>& cells, int N) {
      unordered_map<string, int> map;
      string firstcell = "";
      for (int i = ; i<cells.size(); i++) {
          firstcell += to_string(cells[i]);
      }
      while (N != ) {
         if (map.count(firstcell))
             N %= map[firstcell] - N;
         if(N == ) break;
         string nextstr = "";
         for (int i = ; i < ; i++) {
             nextstr += firstcell[i - ] == firstcell[i + ] ? "" : "";
         }
         nextstr = "" + nextstr + "";
         //cout << nextstr << endl;
         map[firstcell] = N;
         firstcell = nextstr;
         N--;
     }
     vector<int> ret;
     for (int i = ; i<firstcell.size(); i++) {
         if (firstcell[i] == '') ret.push_back();
         else ret.push_back();
     }
     return ret;
 }
 };