0519. Random Flip Matrix

# 519. Random Flip Matrix#

## 题目 #

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.

## 题目大意 #

• 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

## 解题思路 #

• 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

## 代码 #

``````package leetcode

import "math/rand"

type Solution struct {
r     int
c     int
total int
mp    map[int]int
}

func Constructor(m int, n int) Solution {
return Solution{
r:     m,
c:     n,
total: m * n,
mp:    map[int]int{},
}
}

func (this *Solution) Flip() []int {
k := rand.Intn(this.total)
val := k
if v, 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.c
return []int{newR, newC}
}

func (this *Solution) Reset() {
this.total = this.r * this.c
this.mp = map[int]int{}
}
``````

Apr 8, 2023