feat: add solutions to lc problems: No.0292,0293 (#2308)

doocs · Feb 4, 2024 · c33bcc7 · c33bcc7
1 parent 859c150
commit c33bcc7
Show file tree

Hide file tree

Showing 9 changed files with 148 additions and 89 deletions.
diff --git a/solution/0200-0299/0292.Nim Game/README.md b/solution/0200-0299/0292.Nim Game/README.md
@@ -55,19 +55,20 @@
 
 ## 解法
 
-### 方法一：数学推理
+### 方法一：找规律
 
 第一个得到 $4$ 的倍数（即 $n$ 能被 $4$ 整除）的将会输掉比赛。
 
 证明：
 
-1. 当 $n=4$，无论第一个玩家选择 $1/2/3$ 哪个数字，第二个玩家总能选择剩下的数字，**第一个玩家将会输掉比赛**。
-1. 当 $4<n<8$，即 ($n=5,6,7$)，第一个玩家可以相应地将数字减少为 $4$，那么 $4$ 这个死亡数字给到了第二个玩家，第二个玩家将会输掉比赛。
-1. 当 $n=8$，无论第一个玩家选择 $1/2/3$ 哪个数字，都会把 $4<n<8$ 的数字留给第二个，**第一个玩家将会输掉比赛**。
+1. 当 $n \lt 4$ 时，第一个玩家可以直接拿走所有的石头，所以第一个玩家将会赢得比赛。
+1. 当 $n = 4$，无论第一个玩家选择 $1, 2, 3$ 哪个数字，第二个玩家总能选择剩下的数字，所以第一个玩家将会输掉比赛。
+1. 当 $4 \lt n \lt 8$ 时，即 $n = 5, 6, 7$，第一个玩家可以相应地将数字减少为 $4$，那么 $4$ 这个死亡数字给到了第二个玩家，第二个玩家将会输掉比赛。
+1. 当 $n = 8$，无论第一个玩家选择 $1, 2, 3$ 哪个数字，都会把 $4 \lt n \lt 8$ 的数字留给第二个，所以第一个玩家将会输掉比赛。
 1. ...
 1. 依次类推，当玩家拿到 $n$ 这个数字，且 $n$ 能被 $4$ 整除，他将会输掉比赛，否则他将赢得比赛。
 
-时间复杂度 $O(1)$。
+时间复杂度 $O(1)$，空间复杂度 $O(1)$。
 
 <!-- tabs:start -->
 

diff --git a/solution/0200-0299/0292.Nim Game/README_EN.md b/solution/0200-0299/0292.Nim Game/README_EN.md
@@ -51,7 +51,20 @@ In all outcomes, your friend wins.
 
 ## Solutions
 
-### Solution 1
+### Solution 1: Finding the Pattern
+
+The first player who gets a multiple of $4$ (i.e., $n$ can be divided by $4$) will lose the game.
+
+Proof:
+
+1. When $n \lt 4$, the first player can directly take all the stones, so the first player will win the game.
+1. When $n = 4$, no matter whether the first player chooses $1, 2, 3$, the second player can always choose the remaining number, so the first player will lose the game.
+1. When $4 \lt n \lt 8$, i.e., $n = 5, 6, 7$, the first player can correspondingly reduce the number to $4$, then the "death number" $4$ is given to the second player, and the second player will lose the game.
+1. When $n = 8$, no matter whether the first player chooses $1, 2, 3$, it will leave a number between $4 \lt n \lt 8$ to the second player, so the first player will lose the game.
+1. ...
+1. By induction, when a player gets the number $n$, and $n$ can be divided by $4$, he will lose the game, otherwise, he will win the game.
+
+The time complexity is $O(1)$, and the space complexity is $O(1)$.
 
 <!-- tabs:start -->
 

diff --git a/solution/0200-0299/0293.Flip Game/README.md b/solution/0200-0299/0293.Flip Game/README.md
@@ -39,7 +39,13 @@
 
 ## 解法
 
-### 方法一
+### 方法一：遍历 + 模拟
+
+我们遍历字符串，如果当前字符和下一个字符都是 `+`，那么我们就将这两个字符变成 `-`，然后将结果加入到结果数组中，再将这两个字符变回 `+`。
+
+遍历结束后，返回结果数组即可。
+
+时间复杂度 $O(n^2)$，其中 $n$ 是字符串长度。忽略答案数组的空间复杂度，空间复杂度 $O(n)$ 或 $O(1)$。
 
 <!-- tabs:start -->
 
@@ -48,8 +54,8 @@ class Solution:
     def generatePossibleNextMoves(self, currentState: str) -> List[str]:
         s = list(currentState)
         ans = []
-        for i, c in enumerate(s[:-1]):
-            if c == "+" and s[i + 1] == "+":
+        for i, (a, b) in enumerate(pairwise(s)):
+            if a == b == "+":
                 s[i] = s[i + 1] = "-"
                 ans.append("".join(s))
                 s[i] = s[i + 1] = "+"
@@ -59,15 +65,15 @@ class Solution:
 ```java
 class Solution {
     public List<String> generatePossibleNextMoves(String currentState) {
-        char[] cs = currentState.toCharArray();
         List<String> ans = new ArrayList<>();
-        for (int i = 0; i < cs.length - 1; ++i) {
-            if (cs[i] == '+' && cs[i + 1] == '+') {
-                cs[i] = '-';
-                cs[i + 1] = '-';
-                ans.add(String.valueOf(cs));
-                cs[i] = '+';
-                cs[i + 1] = '+';
+        char[] s = currentState.toCharArray();
+        for (int i = 0; i < s.length - 1; ++i) {
+            if (s[i] == '+' && s[i + 1] == '+') {
+                s[i] = '-';
+                s[i + 1] = '-';
+                ans.add(new String(s));
+                s[i] = '+';
+                s[i + 1] = '+';
             }
         }
         return ans;
@@ -78,15 +84,13 @@ class Solution {
 ```cpp
 class Solution {
 public:
-    vector<string> generatePossibleNextMoves(string currentState) {
+    vector<string> generatePossibleNextMoves(string s) {
         vector<string> ans;
-        for (int i = 0; i < currentState.size() - 1; ++i) {
-            if (currentState[i] == '+' && currentState[i + 1] == '+') {
-                currentState[i] = '-';
-                currentState[i + 1] = '-';
-                ans.push_back(currentState);
-                currentState[i] = '+';
-                currentState[i + 1] = '+';
+        for (int i = 0; i < s.size() - 1; ++i) {
+            if (s[i] == '+' && s[i + 1] == '+') {
+                s[i] = s[i + 1] = '-';
+                ans.emplace_back(s);
+                s[i] = s[i + 1] = '+';
             }
         }
         return ans;
@@ -95,17 +99,31 @@ public:
 ```
 
 ```go
-func generatePossibleNextMoves(currentState string) []string {
-	ans := []string{}
-	cs := []byte(currentState)
-	for i, c := range cs[1:] {
-		if c == '+' && cs[i] == '+' {
-			cs[i], cs[i+1] = '-', '-'
-			ans = append(ans, string(cs))
-			cs[i], cs[i+1] = '+', '+'
+func generatePossibleNextMoves(currentState string) (ans []string) {
+	s := []byte(currentState)
+	for i := 0; i < len(s)-1; i++ {
+		if s[i] == '+' && s[i+1] == '+' {
+			s[i], s[i+1] = '-', '-'
+			ans = append(ans, string(s))
+			s[i], s[i+1] = '+', '+'
 		}
 	}
-	return ans
+	return
+}
+```
+
+```ts
+function generatePossibleNextMoves(currentState: string): string[] {
+    const s = currentState.split('');
+    const ans: string[] = [];
+    for (let i = 0; i < s.length - 1; ++i) {
+        if (s[i] === '+' && s[i + 1] === '+') {
+            s[i] = s[i + 1] = '-';
+            ans.push(s.join(''));
+            s[i] = s[i + 1] = '+';
+        }
+    }
+    return ans;
 }
 ```
 

diff --git a/solution/0200-0299/0293.Flip Game/README_EN.md b/solution/0200-0299/0293.Flip Game/README_EN.md
@@ -35,7 +35,13 @@
 
 ## Solutions
 
-### Solution 1
+### Solution 1: Traversal + Simulation
+
+We traverse the string. If the current character and the next character are both `+`, we change these two characters to `-`, add the result to the result array, and then change these two characters back to `+`.
+
+After the traversal ends, we return the result array.
+
+The time complexity is $O(n^2)$, where $n$ is the length of the string. Ignoring the space complexity of the result array, the space complexity is $O(n)$ or $O(1)$.
 
 <!-- tabs:start -->
 
@@ -44,8 +50,8 @@ class Solution:
     def generatePossibleNextMoves(self, currentState: str) -> List[str]:
         s = list(currentState)
         ans = []
-        for i, c in enumerate(s[:-1]):
-            if c == "+" and s[i + 1] == "+":
+        for i, (a, b) in enumerate(pairwise(s)):
+            if a == b == "+":
                 s[i] = s[i + 1] = "-"
                 ans.append("".join(s))
                 s[i] = s[i + 1] = "+"
@@ -55,15 +61,15 @@ class Solution:
 ```java
 class Solution {
     public List<String> generatePossibleNextMoves(String currentState) {
-        char[] cs = currentState.toCharArray();
         List<String> ans = new ArrayList<>();
-        for (int i = 0; i < cs.length - 1; ++i) {
-            if (cs[i] == '+' && cs[i + 1] == '+') {
-                cs[i] = '-';
-                cs[i + 1] = '-';
-                ans.add(String.valueOf(cs));
-                cs[i] = '+';
-                cs[i + 1] = '+';
+        char[] s = currentState.toCharArray();
+        for (int i = 0; i < s.length - 1; ++i) {
+            if (s[i] == '+' && s[i + 1] == '+') {
+                s[i] = '-';
+                s[i + 1] = '-';
+                ans.add(new String(s));
+                s[i] = '+';
+                s[i + 1] = '+';
             }
         }
         return ans;
@@ -74,15 +80,13 @@ class Solution {
 ```cpp
 class Solution {
 public:
-    vector<string> generatePossibleNextMoves(string currentState) {
+    vector<string> generatePossibleNextMoves(string s) {
         vector<string> ans;
-        for (int i = 0; i < currentState.size() - 1; ++i) {
-            if (currentState[i] == '+' && currentState[i + 1] == '+') {
-                currentState[i] = '-';
-                currentState[i + 1] = '-';
-                ans.push_back(currentState);
-                currentState[i] = '+';
-                currentState[i + 1] = '+';
+        for (int i = 0; i < s.size() - 1; ++i) {
+            if (s[i] == '+' && s[i + 1] == '+') {
+                s[i] = s[i + 1] = '-';
+                ans.emplace_back(s);
+                s[i] = s[i + 1] = '+';
             }
         }
         return ans;
@@ -91,17 +95,31 @@ public:
 ```
 
 ```go
-func generatePossibleNextMoves(currentState string) []string {
-	ans := []string{}
-	cs := []byte(currentState)
-	for i, c := range cs[1:] {
-		if c == '+' && cs[i] == '+' {
-			cs[i], cs[i+1] = '-', '-'
-			ans = append(ans, string(cs))
-			cs[i], cs[i+1] = '+', '+'
+func generatePossibleNextMoves(currentState string) (ans []string) {
+	s := []byte(currentState)
+	for i := 0; i < len(s)-1; i++ {
+		if s[i] == '+' && s[i+1] == '+' {
+			s[i], s[i+1] = '-', '-'
+			ans = append(ans, string(s))
+			s[i], s[i+1] = '+', '+'
 		}
 	}
-	return ans
+	return
+}
+```
+
+```ts
+function generatePossibleNextMoves(currentState: string): string[] {
+    const s = currentState.split('');
+    const ans: string[] = [];
+    for (let i = 0; i < s.length - 1; ++i) {
+        if (s[i] === '+' && s[i + 1] === '+') {
+            s[i] = s[i + 1] = '-';
+            ans.push(s.join(''));
+            s[i] = s[i + 1] = '+';
+        }
+    }
+    return ans;
 }
 ```
 

diff --git a/solution/0200-0299/0293.Flip Game/Solution.cpp b/solution/0200-0299/0293.Flip Game/Solution.cpp
@@ -1,14 +1,12 @@
 class Solution {
 public:
-    vector<string> generatePossibleNextMoves(string currentState) {
+    vector<string> generatePossibleNextMoves(string s) {
         vector<string> ans;
-        for (int i = 0; i < currentState.size() - 1; ++i) {
-            if (currentState[i] == '+' && currentState[i + 1] == '+') {
-                currentState[i] = '-';
-                currentState[i + 1] = '-';
-                ans.push_back(currentState);
-                currentState[i] = '+';
-                currentState[i + 1] = '+';
+        for (int i = 0; i < s.size() - 1; ++i) {
+            if (s[i] == '+' && s[i + 1] == '+') {
+                s[i] = s[i + 1] = '-';
+                ans.emplace_back(s);
+                s[i] = s[i + 1] = '+';
             }
         }
         return ans;

diff --git a/solution/0200-0299/0293.Flip Game/Solution.go b/solution/0200-0299/0293.Flip Game/Solution.go
@@ -1,12 +1,11 @@
-func generatePossibleNextMoves(currentState string) []string {
-	ans := []string{}
-	cs := []byte(currentState)
-	for i, c := range cs[1:] {
-		if c == '+' && cs[i] == '+' {
-			cs[i], cs[i+1] = '-', '-'
-			ans = append(ans, string(cs))
-			cs[i], cs[i+1] = '+', '+'
+func generatePossibleNextMoves(currentState string) (ans []string) {
+	s := []byte(currentState)
+	for i := 0; i < len(s)-1; i++ {
+		if s[i] == '+' && s[i+1] == '+' {
+			s[i], s[i+1] = '-', '-'
+			ans = append(ans, string(s))
+			s[i], s[i+1] = '+', '+'
 		}
 	}
-	return ans
+	return
 }
diff --git a/solution/0200-0299/0293.Flip Game/Solution.java b/solution/0200-0299/0293.Flip Game/Solution.java
@@ -1,14 +1,14 @@
 class Solution {
     public List<String> generatePossibleNextMoves(String currentState) {
-        char[] cs = currentState.toCharArray();
         List<String> ans = new ArrayList<>();
-        for (int i = 0; i < cs.length - 1; ++i) {
-            if (cs[i] == '+' && cs[i + 1] == '+') {
-                cs[i] = '-';
-                cs[i + 1] = '-';
-                ans.add(String.valueOf(cs));
-                cs[i] = '+';
-                cs[i + 1] = '+';
+        char[] s = currentState.toCharArray();
+        for (int i = 0; i < s.length - 1; ++i) {
+            if (s[i] == '+' && s[i + 1] == '+') {
+                s[i] = '-';
+                s[i + 1] = '-';
+                ans.add(new String(s));
+                s[i] = '+';
+                s[i + 1] = '+';
             }
         }
         return ans;

diff --git a/solution/0200-0299/0293.Flip Game/Solution.py b/solution/0200-0299/0293.Flip Game/Solution.py
@@ -2,8 +2,8 @@ class Solution:
     def generatePossibleNextMoves(self, currentState: str) -> List[str]:
         s = list(currentState)
         ans = []
-        for i, c in enumerate(s[:-1]):
-            if c == "+" and s[i + 1] == "+":
+        for i, (a, b) in enumerate(pairwise(s)):
+            if a == b == "+":
                 s[i] = s[i + 1] = "-"
                 ans.append("".join(s))
                 s[i] = s[i + 1] = "+"

diff --git a/solution/0200-0299/0293.Flip Game/Solution.ts b/solution/0200-0299/0293.Flip Game/Solution.ts
@@ -0,0 +1,12 @@
+function generatePossibleNextMoves(currentState: string): string[] {
+    const s = currentState.split('');
+    const ans: string[] = [];
+    for (let i = 0; i < s.length - 1; ++i) {
+        if (s[i] === '+' && s[i + 1] === '+') {
+            s[i] = s[i + 1] = '-';
+            ans.push(s.join(''));
+            s[i] = s[i + 1] = '+';
+        }
+    }
+    return ans;
+}