957. Prison Cells After N Days - medium
문제
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.)
제한사항
- cells.length == 8
- cells[i] is in {0, 1}
- 1 <= N <= 10^9
입출력 예
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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]
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Example 2:
Input: cells = [1,0,0,1,0,0,1,0], N = 1000000000
Output: [0,0,1,1,1,1,1,0]
풀이
- hash
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class Solution {
public:
vector<int> prisonAfterNDays(vector<int>& cells, int N) {
std::vector<std::vector<int>> item;
// 순환되는 횟수를 찾을때 까지 계산한 모든 cells를 저장
while (N--) {
std::vector<int> temp(8, 0);
for (int i = 1; i < 7; i++) {
temp[i] = (cells[i - 1] == cells[i + 1]) ? 1 : 0;
}
// 순환되는 것을 찾으면 인덱스 계산
if (item.size() && item.front() == temp) {
return item[N % item.size()];
} else {
item.emplace_back(temp);
}
cells = temp;
}
return cells;
}
};