Add solution and test-cases for problem 957

Author: 0xff-dev

leetcode/901-1000/0957.Prison-Cells-After-N-Days/README.md

@@ -1,28 +1,41 @@
 # [957.Prison Cells After N Days][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
+There are 8 prison cells in a row and each cell is either occupied or vacant.
+
+Each day, whether the cell is occupied or vacant changes according to the following rules:
+
+- If a cell has two adjacent neighbors that are both occupied or both vacant, then the cell becomes occupied.
+- Otherwise, it becomes vacant.
+
+**Note** that because the prison is a row, the first and the last cells in the row can't have two adjacent neighbors.
+
+You are given an integer array `cells` where `cells[i] == 1` if the `ith` cell is occupied and `cells[i] == 0` if the `ith` cell is vacant, and you are given an integer n.
+
+Return the state of the prison after `n` days (i.e., n such changes described above).
 
 **Example 1:**
 
 ```
-Input: a = "11", b = "1"
-Output: "100"
+Input: cells = [0,1,0,1,1,0,0,1], n = 7
+Output: [0,0,1,1,0,0,0,0]
+Explanation: The following table summarizes the state of the prison on each day:
+Day 0: [0, 1, 0, 1, 1, 0, 0, 1]
+Day 1: [0, 1, 1, 0, 0, 0, 0, 0]
+Day 2: [0, 0, 0, 0, 1, 1, 1, 0]
+Day 3: [0, 1, 1, 0, 0, 1, 0, 0]
+Day 4: [0, 0, 0, 0, 0, 1, 0, 0]
+Day 5: [0, 1, 1, 1, 0, 1, 0, 0]
+Day 6: [0, 0, 1, 0, 1, 1, 0, 0]
+Day 7: [0, 0, 1, 1, 0, 0, 0, 0]
 ```
 
-## 题意
-> ...
-
-## 题解
+**Example 2:**
 
-### 思路1
-> ...
-Prison Cells After N Days
-```go
 ```
-
+Input: cells = [1,0,0,1,0,0,1,0], n = 1000000000
+Output: [0,0,1,1,1,1,1,0]
+```
 
 ## 结语
 

leetcode/901-1000/0957.Prison-Cells-After-N-Days/Solution.go

@@ -1,5 +1,45 @@
 package Solution
 
-func Solution(x bool) bool {
-	return x
+func Solution(cells []int, n int) []int {
+	start := [8]int{}
+	for i := range cells {
+		start[i] = cells[i]
+	}
+	days := [][8]int{start}
+	day := 0
+	cache := map[[8]int]int{
+		start: day,
+	}
+
+	var convert func([8]int) [8]int
+	convert = func(now [8]int) [8]int {
+		var res [8]int
+		res[0], res[7] = 0, 0
+		for i := 1; i < 7; i++ {
+			if now[i-1] == now[i+1] {
+				res[i] = 1
+			}
+		}
+		return res
+	}
+	var r [8]int
+	var loopStart, loopLen int
+	for {
+		day++
+		r = convert(start)
+		if day == n {
+			return r[:]
+		}
+		if d, ok := cache[r]; ok {
+			loopStart = d
+			loopLen = day - d
+			break
+		}
+		cache[r] = day
+		start = r
+		days = append(days, r)
+	}
+	target := n - loopStart
+	target %= loopLen
+	return days[loopStart+target][:]
 }

leetcode/901-1000/0957.Prison-Cells-After-N-Days/Solution_test.go

@@ -10,30 +10,30 @@ func TestSolution(t *testing.T) {
 	//	测试用例
 	cases := []struct {
 		name   string
-		inputs bool
-		expect bool
+		cells  []int
+		n      int
+		expect []int
 	}{
-		{"TestCase", true, true},
-		{"TestCase", true, true},
-		{"TestCase", false, false},
+		{"TestCase1", []int{0, 1, 0, 1, 1, 0, 0, 1}, 7, []int{0, 0, 1, 1, 0, 0, 0, 0}},
+		{"TestCase2", []int{1, 0, 0, 1, 0, 0, 1, 0}, 1000000000, []int{0, 0, 1, 1, 1, 1, 1, 0}},
 	}
 
 	//	开始测试
 	for i, c := range cases {
 		t.Run(c.name+" "+strconv.Itoa(i), func(t *testing.T) {
-			got := Solution(c.inputs)
+			got := Solution(c.cells, c.n)
 			if !reflect.DeepEqual(got, c.expect) {
-				t.Fatalf("expected: %v, but got: %v, with inputs: %v",
-					c.expect, got, c.inputs)
+				t.Fatalf("expected: %v, but got: %v, with inputs: %v %v",
+					c.expect, got, c.cells, c.n)
 			}
 		})
 	}
 }
 
-//	压力测试
+// 压力测试
 func BenchmarkSolution(b *testing.B) {
 }
 
-//	使用案列
+// 使用案列
 func ExampleSolution() {
 }