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98 lines
3.2 KiB
Markdown
98 lines
3.2 KiB
Markdown
# [519. Random Flip Matrix](https://leetcode.com/problems/random-flip-matrix/)
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## 题目
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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.
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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.
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Implement the Solution class:
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- Solution(int m, int n) Initializes the object with the size of the binary matrix m and n.
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- int[] flip() Returns a random index [i, j] of the matrix where matrix[i][j] == 0 and flips it to 1.
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- void reset() Resets all the values of the matrix to be 0.
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**Example 1**:
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Input
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["Solution", "flip", "flip", "flip", "reset", "flip"]
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[[3, 1], [], [], [], [], []]
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Output
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[null, [1, 0], [2, 0], [0, 0], null, [2, 0]]
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Explanation
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Solution solution = new Solution(3, 1);
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solution.flip(); // return [1, 0], [0,0], [1,0], and [2,0] should be equally likely to be returned.
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solution.flip(); // return [2, 0], Since [1,0] was returned, [2,0] and [0,0]
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solution.flip(); // return [0, 0], Based on the previously returned indices, only [0,0] can be returned.
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solution.reset(); // All the values are reset to 0 and can be returned.
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solution.flip(); // return [2, 0], [0,0], [1,0], and [2,0] should be equally likely to be returned.
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**Constraints:**
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- 1 <= m, n <= 10000
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- There will be at least one free cell for each call to flip.
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- At most 1000 calls will be made to flip and reset.
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## 题目大意
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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。
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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。
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实现 Solution 类：
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- Solution(int m, int n) 使用二元矩阵的大小 m 和 n 初始化该对象
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- 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
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- void reset() 将矩阵中所有的值重置为 0
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## 解题思路
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- 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
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- 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
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## 代码
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```go
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package leetcode
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import "math/rand"
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type Solution struct {
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r int
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c int
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total int
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mp map[int]int
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}
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func Constructor(m int, n int) Solution {
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return Solution{
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r: m,
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c: n,
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total: m * n,
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mp: map[int]int{},
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}
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}
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func (this *Solution) Flip() []int {
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k := rand.Intn(this.total)
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val := k
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if v, ok := this.mp[k]; ok {
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val = v
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}
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if _, ok := this.mp[this.total-1]; ok {
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this.mp[k] = this.mp[this.total-1]
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} else {
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this.mp[k] = this.total - 1
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}
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delete(this.mp, this.total - 1)
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this.total--
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newR, newC := val/this.c, val%this.c
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return []int{newR, newC}
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}
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func (this *Solution) Reset() {
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this.total = this.r * this.c
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this.mp = map[int]int{}
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}
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``` |