https://leetcode.com/problems/random-flip-matrix/
You are given the number of rows
n_rowsand number of columnsn_colsof a 2D binary matrix where all values are initially 0. Write a functionflipwhich chooses a 0 value uniformly at random, changes it to 1, and then returns the position[row.id, col.id]of that value. Also, write a functionresetwhich sets all values back to 0. Try to minimize the number of calls to system's Math.random() and optimize the time and space complexity.Note:
1 <= n_rows, n_cols <= 100000 <= row.id < n_rowsand0 <= col.id < n_colsflipwill not be called when the matrix has no 0 values left.- the total number of calls to
flipandresetwill not exceed 1000.Example 1:
Input: ["Solution","flip","flip","flip","flip"] [[2,3],[],[],[],[]] Output: [null,[0,1],[1,2],[1,0],[1,1]]Example 2:
Input: ["Solution","flip","flip","reset","flip"] [[1,2],[],[],[],[]] Output: [null,[0,0],[0,1],null,[0,0]]Explanation of Input Syntax:
The input is two lists: the subroutines called and their arguments.
Solution's constructor has two arguments,n_rowsandn_cols.flipandresethave no arguments. Arguments are always wrapped with a list, even if there aren't any.
流随机会超时,因为调用太多次rand()
使用逻辑位置与物理位置的重映射
class Solution {
public:
int row, col;
int total;
unordered_map<int, int> M;
Solution(int n_rows, int n_cols) {
row = n_rows;
col = n_cols;
total = row*col;
}
vector<int> flip() {
int logic_id = rand() % total;
int physic_id = logic_id;
if(M.find(logic_id) != M.end()) physic_id = M[logic_id];
M[logic_id] = (M.find(total-1) == M.end()) ? total-1 : M[total-1];
--total;
return {physic_id / col, physic_id % col};
}
void reset() {
total = row*col;
M.clear();
}
};
/**
* Your Solution object will be instantiated and called as such:
* Solution* obj = new Solution(n_rows, n_cols);
* vector<int> param_1 = obj->flip();
* obj->reset();
*/
LeetCode随机翻转矩阵问题解法

博客围绕LeetCode上的随机翻转矩阵问题展开,要求编写函数随机选0值置为1并返回位置,还有函数将值全置为0,需减少Math.random()调用并优化时空复杂度。指出流随机会超时,可使用逻辑与物理位置重映射解决。
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