Add solution 0576

leetcode/0576.Out-of-Boundary-Paths/576. Out of Boundary Paths.go

@@ -0,0 +1,43 @@
+package leetcode
+
+var dir = [][]int{
+	{-1, 0},
+	{0, 1},
+	{1, 0},
+	{0, -1},
+}
+
+func findPaths(m int, n int, maxMove int, startRow int, startColumn int) int {
+	visited := make([][][]int, m)
+	for i := range visited {
+		visited[i] = make([][]int, n)
+		for j := range visited[i] {
+			visited[i][j] = make([]int, maxMove+1)
+			for l := range visited[i][j] {
+				visited[i][j][l] = -1
+			}
+		}
+	}
+	return dfs(startRow, startColumn, maxMove, m, n, visited)
+}
+
+func dfs(x, y, maxMove, m, n int, visited [][][]int) int {
+	if x < 0 || x >= m || y < 0 || y >= n {
+		return 1
+	}
+	if maxMove == 0 {
+		visited[x][y][maxMove] = 0
+		return 0
+	}
+	if visited[x][y][maxMove] >= 0 {
+		return visited[x][y][maxMove]
+	}
+	res := 0
+	for i := 0; i < 4; i++ {
+		nx := x + dir[i][0]
+		ny := y + dir[i][1]
+		res += (dfs(nx, ny, maxMove-1, m, n, visited) % 1000000007)
+	}
+	visited[x][y][maxMove] = res % 1000000007
+	return visited[x][y][maxMove]
+}

leetcode/0576.Out-of-Boundary-Paths/576. Out of Boundary Paths_test.go

@@ -0,0 +1,51 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question576 struct {
+	para576
+	ans576
+}
+
+// para 是参数
+// one 代表第一个参数
+type para576 struct {
+	m           int
+	n           int
+	maxMove     int
+	startRow    int
+	startColumn int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans576 struct {
+	one int
+}
+
+func Test_Problem576(t *testing.T) {
+
+	qs := []question576{
+
+		{
+			para576{2, 2, 2, 0, 0},
+			ans576{6},
+		},
+
+		{
+			para576{1, 3, 3, 0, 1},
+			ans576{12},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 576------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans576, q.para576
+		fmt.Printf("【input】:%v       【output】:%v\n", p, findPaths(p.m, p.n, p.maxMove, p.startRow, p.startColumn))
+	}
+	fmt.Printf("\n\n\n")
+}

leetcode/0576.Out-of-Boundary-Paths/README.md

@@ -0,0 +1,89 @@
+# [576. Out of Boundary Paths](https://leetcode.com/problems/out-of-boundary-paths/)
+
+
+## 题目
+
+There is an `m x n` grid with a ball. The ball is initially at the position `[startRow, startColumn]`. You are allowed to move the ball to one of the four adjacent four cells in the grid (possibly out of the grid crossing the grid boundary). You can apply **at most** `maxMove` moves to the ball.
+
+Given the five integers `m`, `n`, `maxMove`, `startRow`, `startColumn`, return the number of paths to move the ball out of the grid boundary. Since the answer can be very large, return it **modulo** `109 + 7`.
+
+**Example 1:**
+
+![https://assets.leetcode.com/uploads/2021/04/28/out_of_boundary_paths_1.png](https://assets.leetcode.com/uploads/2021/04/28/out_of_boundary_paths_1.png)
+
+```
+Input: m = 2, n = 2, maxMove = 2, startRow = 0, startColumn = 0
+Output: 6
+```
+
+**Example 2:**
+
+![https://assets.leetcode.com/uploads/2021/04/28/out_of_boundary_paths_2.png](https://assets.leetcode.com/uploads/2021/04/28/out_of_boundary_paths_2.png)
+
+```
+Input: m = 1, n = 3, maxMove = 3, startRow = 0, startColumn = 1
+Output: 12
+```
+
+**Constraints:**
+
+- `1 <= m, n <= 50`
+- `0 <= maxMove <= 50`
+- `0 <= startRow <= m`
+- `0 <= startColumn <= n`
+
+## 题目大意
+
+给定一个 m × n 的网格和一个球。球的起始坐标为 (i,j) ，你可以将球移到相邻的单元格内，或者往上、下、左、右四个方向上移动使球穿过网格边界。但是，你最多可以移动 N 次。找出可以将球移出边界的路径数量。答案可能非常大，返回 结果 mod 109 + 7 的值。
+
+## 解题思路
+
+- 单纯暴力的思路，在球的每个方向都遍历一步，直到移动步数用完。这样暴力搜索，解空间是 4^n 。优化思路便是增加记忆化。用三维数组记录位置坐标和步数，对应的出边界的路径数量。加上记忆化以后的深搜解法 runtime beats 100% 了。
+
+## 代码
+
+```go
+package leetcode
+
+var dir = [][]int{
+	{-1, 0},
+	{0, 1},
+	{1, 0},
+	{0, -1},
+}
+
+func findPaths(m int, n int, maxMove int, startRow int, startColumn int) int {
+	visited := make([][][]int, m)
+	for i := range visited {
+		visited[i] = make([][]int, n)
+		for j := range visited[i] {
+			visited[i][j] = make([]int, maxMove+1)
+			for l := range visited[i][j] {
+				visited[i][j][l] = -1
+			}
+		}
+	}
+	return dfs(startRow, startColumn, maxMove, m, n, visited)
+}
+
+func dfs(x, y, maxMove, m, n int, visited [][][]int) int {
+	if x < 0 || x >= m || y < 0 || y >= n {
+		return 1
+	}
+	if maxMove == 0 {
+		visited[x][y][maxMove] = 0
+		return 0
+	}
+	if visited[x][y][maxMove] >= 0 {
+		return visited[x][y][maxMove]
+	}
+	res := 0
+	for i := 0; i < 4; i++ {
+		nx := x + dir[i][0]
+		ny := y + dir[i][1]
+		res += (dfs(nx, ny, maxMove-1, m, n, visited) % 1000000007)
+	}
+	visited[x][y][maxMove] = res % 1000000007
+	return visited[x][y][maxMove]
+}
+```