There is an m x n binary grid matrix with all the values set 0 initially. Design an algorithm to randomly pick an index (i, j) where matrix[i][j] == 0 and flips it to 1. All the indices (i, j) where matrix[i][j] == 0 should be equally likely to be returned.

Optimize your algorithm to minimize the number of calls made to the built-in random function of your language and optimize the time and space complexity.

Implement the Solution class:

• Solution(int m, int n) Initializes the object with the size of the binary matrix m and n.
• int[] flip() Returns a random index [i, j] of the matrix where matrix[i][j] == 0 and flips it to 1.
• void reset() Resets all the values of the matrix to be 0.

Example 1:

Input ["Solution", "flip", "flip", "flip", "reset", "flip"] [[3, 1], [], [], [], [], []] Output [null, [1, 0], [2, 0], [0, 0], null, [2, 0]] Explanation Solution solution = new Solution(3, 1); solution.flip(); // return [1, 0], [0,0], [1,0], and [2,0] should be equally likely to be returned. solution.flip(); // return [2, 0], Since [1,0] was returned, [2,0] and [0,0] solution.flip(); // return [0, 0], Based on the previously returned indices, only [0,0] can be returned. solution.reset(); // All the values are reset to 0 and can be returned. solution.flip(); // return [2, 0], [0,0], [1,0], and [2,0] should be equally likely to be returned. 

Constraints:

• 1 <= m, n <= 10000
• There will be at least one free cell for each call to flip.
• At most 1000 calls will be made to flip and reset.

给你一个 m x n 的二元矩阵 matrix ，且所有值被初始化为 0 。请你设计一个算法，随机选取一个满足 matrix[i][j] == 0 的下标 (i, j) ，并将它的值变为 1 。所有满足 matrix[i][j] == 0 的下标 (i, j) 被选取的概率应当均等。

尽量最少调用内置的随机函数，并且优化时间和空间复杂度。

实现 Solution 类：

• Solution(int m, int n) 使用二元矩阵的大小 m 和 n 初始化该对象
• int[] flip() 返回一个满足 matrix[i][j] == 0 的随机下标 [i, j] ，并将其对应格子中的值变为 1
• void reset() 将矩阵中所有的值重置为 0

• 二维矩阵利用哈希表转换为一维,每次随机选择一维中的任意一个元素，然后与最后一个元素交换，一维元素的总个数减一
• 哈希表中默认的映射为x->x, 然后将不满足这个映射的特殊键值对存入哈希表

package leetcode import"math/rand"typeSolutionstruct { rintcinttotalintmpmap[int]int } funcConstructor(mint, nint) Solution { returnSolution{ r: m, c: n, total: m*n, mp: map[int]int{}, } } func (this*Solution) Flip() []int { k:=rand.Intn(this.total) val:=kifv, ok:=this.mp[k]; ok { val=v } if_, ok:=this.mp[this.total-1]; ok { this.mp[k] =this.mp[this.total-1] } else { this.mp[k] =this.total-1 } delete(this.mp, this.total-1) this.total--newR, newC:=val/this.c, val%this.creturn []int{newR, newC} } func (this*Solution) Reset() { this.total=this.r*this.cthis.mp=map[int]int{} }