Merge pull request #862 from 0xff-dev/994

Add solution and test-cases for problem 994

Author: 6boris

leetcode/901-1000/0994.Rotting-Oranges/README.md

@@ -1,28 +1,40 @@
 # [994.Rotting Oranges][title]
 
-> [!WARNING|style:flat]
-> This question is temporarily unanswered if you have good ideas. Welcome to [Create Pull Request PR](https://github.com/kylesliu/awesome-golang-algorithm)
-
 ## Description
+You are given an `m x n` grid where each cell can have one of three values:
+
+- `0` representing an empty cell,
+- `1` representing a fresh orange, or
+- `2` representing a rotten orange.
+
+Every minute, any fresh orange that is **4-directionally adjacent** to a rotten orange becomes rotten.
+
+Return the minimum number of minutes that must elapse until no cell has a fresh orange. If this is impossible, return `-1`.
+
+**Example 1:**  
 
-**Example 1:**
+![1](./oranges.png)
 
 ```
-Input: a = "11", b = "1"
-Output: "100"
+Input: grid = [[2,1,1],[1,1,0],[0,1,1]]
+Output: 4
 ```
 
-## 题意
-> ...
+**Example 2:**
 
-## 题解
-
-### 思路1
-> ...
-Rotting Oranges
-```go
 ```
+Input: grid = [[2,1,1],[0,1,1],[1,0,1]]
+Output: -1
+Explanation: The orange in the bottom left corner (row 2, column 0) is never rotten, because rotting only happens 4-directionally.
+```
+
+**Example 3:**
 
+```
+Input: grid = [[0,2]]
+Output: 0
+Explanation: Since there are already no fresh oranges at minute 0, the answer is just 0.
+```
 
 ## 结语
 

leetcode/901-1000/0994.Rotting-Oranges/Solution.go

@@ -1,5 +1,47 @@
 package Solution
 
-func Solution(x bool) bool {
-	return x
+func Solution(grid [][]int) int {
+	used := map[[2]int]struct{}{}
+	rows, cols := len(grid), len(grid[0])
+	freshOranges := 0
+	queue := [][2]int{}
+	for r := 0; r < rows; r++ {
+		for c := 0; c < cols; c++ {
+			if grid[r][c] == 2 {
+				queue = append(queue, [2]int{r, c})
+				used[[2]int{r, c}] = struct{}{}
+			}
+			if grid[r][c] == 1 {
+				freshOranges++
+			}
+		}
+	}
+	var dirs = [][2]int{
+		{-1, 0}, {1, 0}, {0, 1}, {0, -1},
+	}
+	steps := 0
+	for len(queue) > 0 {
+		nq := make([][2]int, 0)
+		for _, cur := range queue {
+			for _, dir := range dirs {
+				nx, ny := cur[0]+dir[0], cur[1]+dir[1]
+				if nx >= 0 && nx < rows && ny >= 0 && ny < cols && grid[nx][ny] == 1 {
+					key := [2]int{nx, ny}
+					if _, ok := used[key]; !ok {
+						nq = append(nq, key)
+						used[key] = struct{}{}
+					}
+				}
+			}
+		}
+		freshOranges -= len(nq)
+		if len(nq) != 0 {
+			steps++
+		}
+		queue = nq
+	}
+	if freshOranges > 0 {
+		return -1
+	}
+	return steps
 }

leetcode/901-1000/0994.Rotting-Oranges/Solution_test.go

@@ -10,12 +10,16 @@ func TestSolution(t *testing.T) {
 	//	测试用例
 	cases := []struct {
 		name   string
-		inputs bool
-		expect bool
+		inputs [][]int
+		expect int
 	}{
-		{"TestCase", true, true},
-		{"TestCase", true, true},
-		{"TestCase", false, false},
+		{"TestCase1", [][]int{
+			{2, 1, 1}, {1, 1, 0}, {0, 1, 1},
+		}, 4},
+		{"TestCase2", [][]int{
+			{2, 1, 1}, {0, 1, 1}, {1, 0, 1},
+		}, -1},
+		{"TestCase3", [][]int{{0, 2}}, 0},
 	}
 
 	//	开始测试
@@ -30,10 +34,10 @@ func TestSolution(t *testing.T) {
 	}
 }
 
-//	压力测试
+// 压力测试
 func BenchmarkSolution(b *testing.B) {
 }
 
-//	使用案列
+// 使用案列
 func ExampleSolution() {
 }