Merge pull request #202 from gostool/leetcode0519

Leetcode0519

leetcode/0519.Random-Flip-Matrix/519.Random Flip Matrix.go

@@ -0,0 +1,43 @@
+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{}
+}

leetcode/0519.Random-Flip-Matrix/519.Random Flip Matrix_test.go

@@ -0,0 +1,52 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question519 struct {
+	para519
+	ans519
+}
+
+// para 是参数
+type para519 struct {
+	para []string
+	val  [][]int
+}
+
+// ans 是答案
+type ans519 struct {
+	ans [][]int
+}
+
+func Test_Problem519(t *testing.T) {
+
+	qs := []question519{
+
+		{
+			para519{[]string{"Solution", "flip", "flip", "flip", "reset", "flip"}, [][]int{{3, 1}, {}, {}, {}, {}, {}}},
+			ans519{[][]int{nil, {1, 0}, {2, 0}, {0, 0}, nil, {2, 0}}},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 519------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans519, q.para519
+		sol := Constructor(0, 0)
+		for _, v := range p.para {
+			if v == "Solution" {
+				sol = Constructor(q.val[0][0], q.val[0][1])
+				fmt.Printf("【input】:%v    【output】:%v\n", v, nil)
+			} else if v == "flip" {
+				fmt.Printf("【input】:%v    【output】:%v\n", v, sol.Flip())
+			} else {
+				sol.Reset()
+				fmt.Printf("【input】:%v    【output】:%v\n", v, nil)
+			}
+		}
+	}
+	fmt.Printf("\n\n\n")
+}

leetcode/0519.Random-Flip-Matrix/README.md

@@ -0,0 +1,98 @@
+# [519. Random Flip Matrix](https://leetcode-cn.com/problems/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
+
+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
+
+## 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
+
+```go
+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{}
+}
+```