LC 957. Prison Cells After N Days
2024-09-05 09:13:10
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;
}
};
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