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。
1 class Solution { 2 public: 3 vector<int> prisonAfterNDays(vector<int>& cells, int N) { 4 unordered_map<string, int> map; 5 string firstcell = ""; 6 for (int i = 0; i<cells.size(); i++) { 7 firstcell += to_string(cells[i]); 8 } 9 while (N != 0) { 10 if (map.count(firstcell)) 11 N %= map[firstcell] - N; 12 if(N == 0) break; 13 string nextstr = ""; 14 for (int i = 1; i < 7; i++) { 15 nextstr += firstcell[i - 1] == firstcell[i + 1] ? "1" : "0"; 16 } 17 nextstr = "0" + nextstr + "0"; 18 //cout << nextstr << endl; 19 map[firstcell] = N; 20 firstcell = nextstr; 21 N--; 22 } 23 vector<int> ret; 24 for (int i = 0; i<firstcell.size(); i++) { 25 if (firstcell[i] == '0') ret.push_back(0); 26 else ret.push_back(1); 27 } 28 return ret; 29 } 30 };