题中给出一个 n_rows
行 n_cols
列的二维矩阵,且所有值被初始化为 0。要求编写一个 flip
函数,均匀随机的将矩阵中的 0 变为 1,并返回该值的位置下标 [row_id,col_id]
;同样编写一个 reset
函数,将所有的值都重新置为 0。尽量最少调用随机函数 Math.random(),并且优化时间和空间复杂度。
注意:
- 1 <= n_rows, n_cols <= 10000
- 0 <= row.id < n_rows 并且 0 <= col.id < n_cols
- 当矩阵中没有值为 0 时,不可以调用 flip 函数
- 调用 flip 和 reset 函数的次数加起来不会超过 1000 次
示例 1:
输入: ["Solution","flip","flip","flip","flip"] [[2,3],[],[],[],[]] 输出: [null,[0,1],[1,2],[1,0],[1,1]]
示例 2:
输入: ["Solution","flip","flip","reset","flip"] [[1,2],[],[],[],[]] 输出: [null,[0,0],[0,1],null,[0,0]]
输入语法解释:
输入包含两个列表:被调用的子程序和他们的参数。Solution
的构造函数有两个参数,分别为 n_rows
和 n_cols
。flip
和 reset
没有参数,参数总会以列表形式给出,哪怕该列表为空
class Solution:
def __init__(self, m: int, n: int):
self.m = m
self.n = n
self.total = m * n
self.mp = {}
def flip(self) -> List[int]:
self.total -= 1
x = random.randint(0, self.total)
idx = self.mp.get(x, x)
self.mp[x] = self.mp.get(self.total, self.total)
return [idx // self.n, idx % self.n]
def reset(self) -> None:
self.total = self.m * self.n
self.mp.clear()
# Your Solution object will be instantiated and called as such:
# obj = Solution(m, n)
# param_1 = obj.flip()
# obj.reset()
class Solution {
private int m;
private int n;
private int total;
private Random rand = new Random();
private Map<Integer, Integer> mp = new HashMap<>();
public Solution(int m, int n) {
this.m = m;
this.n = n;
this.total = m * n;
}
public int[] flip() {
int x = rand.nextInt(total--);
int idx = mp.getOrDefault(x, x);
mp.put(x, mp.getOrDefault(total, total));
return new int[]{idx / n, idx % n};
}
public void reset() {
total = m * n;
mp.clear();
}
}
/**
* Your Solution object will be instantiated and called as such:
* Solution obj = new Solution(m, n);
* int[] param_1 = obj.flip();
* obj.reset();
*/