diff --git a/problems/0001.两数之和.md b/problems/0001.两数之和.md
index 459a66ad..6969c2e2 100644
--- a/problems/0001.两数之和.md
+++ b/problems/0001.两数之和.md
@@ -274,6 +274,30 @@ class Solution {
}
}
```
+
+Scala:
+```scala
+object Solution {
+ // 导入包
+ import scala.collection.mutable
+ def twoSum(nums: Array[Int], target: Int): Array[Int] = {
+ // key存储值，value存储下标
+ val map = new mutable.HashMap[Int, Int]()
+ for (i <- nums.indices) {
+ val tmp = target - nums(i) // 计算差值
+ // 如果这个差值存在于map，则说明找到了结果
+ if (map.contains(tmp)) {
+ return Array(map.get(tmp).get, i)
+ }
+ // 如果不包含把当前值与其下标放到map
+ map.put(nums(i), i)
+ }
+ // 如果没有找到直接返回一个空的数组，return关键字可以省略
+ new Array[Int](2)
+ }
+}
+```
+
C#:
```csharp
public class Solution {
@@ -293,5 +317,6 @@ public class Solution {
}
```
+
-----------------------
diff --git a/problems/0018.四数之和.md b/problems/0018.四数之和.md
index ee70cb69..6cbd40c2 100644
--- a/problems/0018.四数之和.md
+++ b/problems/0018.四数之和.md
@@ -522,6 +522,49 @@ public class Solution
}
}
```
-
+Scala:
+```scala
+object Solution {
+ // 导包
+ import scala.collection.mutable.ListBuffer
+ import scala.util.control.Breaks.{break, breakable}
+ def fourSum(nums: Array[Int], target: Int): List[List[Int]] = {
+ val res = ListBuffer[List[Int]]()
+ val nums_tmp = nums.sorted // 先排序
+ for (i <- nums_tmp.indices) {
+ breakable {
+ if (i > 0 && nums_tmp(i) == nums_tmp(i - 1)) {
+ break // 如果该值和上次的值相同，跳过本次循环，相当于continue
+ } else {
+ for (j <- i + 1 until nums_tmp.length) {
+ breakable {
+ if (j > i + 1 && nums_tmp(j) == nums_tmp(j - 1)) {
+ break // 同上
+ } else {
+ // 双指针
+ var (left, right) = (j + 1, nums_tmp.length - 1)
+ while (left < right) {
+ var sum = nums_tmp(i) + nums_tmp(j) + nums_tmp(left) + nums_tmp(right)
+ if (sum == target) {
+ // 满足要求，直接加入到集合里面去
+ res += List(nums_tmp(i), nums_tmp(j), nums_tmp(left), nums_tmp(right))
+ while (left < right && nums_tmp(left) == nums_tmp(left + 1)) left += 1
+ while (left < right && nums_tmp(right) == nums_tmp(right - 1)) right -= 1
+ left += 1
+ right -= 1
+ } else if (sum < target) left += 1
+ else right -= 1
+ }
+ }
+ }
+ }
+ }
+ }
+ }
+ // 最终返回的res要转换为List，return关键字可以省略
+ res.toList
+ }
+}
+```
-----------------------
diff --git a/problems/0102.二叉树的层序遍历.md b/problems/0102.二叉树的层序遍历.md
index 5f69f53d..1743243d 100644
--- a/problems/0102.二叉树的层序遍历.md
+++ b/problems/0102.二叉树的层序遍历.md
@@ -319,6 +319,36 @@ func levelOrder(_ root: TreeNode?) -> [[Int]] {
}
```
+Rust:
+
+```rust
+pub fn level_order(root: Option>>) -> Vec> {
+ let mut ans = Vec::new();
+ let mut stack = Vec::new();
+ if root.is_none(){
+ return ans;
+ }
+ stack.push(root.unwrap());
+ while stack.is_empty()!= true{
+ let num = stack.len();
+ let mut level = Vec::new();
+ for _i in 0..num{
+ let tmp = stack.remove(0);
+ level.push(tmp.borrow_mut().val);
+ if tmp.borrow_mut().left.is_some(){
+ stack.push(tmp.borrow_mut().left.take().unwrap());
+ }
+ if tmp.borrow_mut().right.is_some(){
+ stack.push(tmp.borrow_mut().right.take().unwrap());
+ }
+ }
+ ans.push(level);
+ }
+ ans
+}
+```
+
+
**此时我们就掌握了二叉树的层序遍历了，那么如下九道力扣上的题目，只需要修改模板的两三行代码（不能再多了），便可打倒！**
@@ -548,6 +578,35 @@ func levelOrderBottom(_ root: TreeNode?) -> [[Int]] {
}
```
+Rust:
+
+```rust
+pub fn level_order(root: Option>>) -> Vec> {
+ let mut ans = Vec::new();
+ let mut stack = Vec::new();
+ if root.is_none(){
+ return ans;
+ }
+ stack.push(root.unwrap());
+ while stack.is_empty()!= true{
+ let num = stack.len();
+ let mut level = Vec::new();
+ for _i in 0..num{
+ let tmp = stack.remove(0);
+ level.push(tmp.borrow_mut().val);
+ if tmp.borrow_mut().left.is_some(){
+ stack.push(tmp.borrow_mut().left.take().unwrap());
+ }
+ if tmp.borrow_mut().right.is_some(){
+ stack.push(tmp.borrow_mut().right.take().unwrap());
+ }
+ }
+ ans.push(level);
+ }
+ ans
+}
+```
+
# 199.二叉树的右视图
[力扣题目链接](https://leetcode-cn.com/problems/binary-tree-right-side-view/)
diff --git a/problems/0104.二叉树的最大深度.md b/problems/0104.二叉树的最大深度.md
index 2229a854..65a155fb 100644
--- a/problems/0104.二叉树的最大深度.md
+++ b/problems/0104.二叉树的最大深度.md
@@ -192,6 +192,33 @@ public:
};
```
+rust:
+```rust
+impl Solution {
+ pub fn max_depth(root: Option>>) -> i32 {
+ if root.is_none(){
+ return 0;
+ }
+ let mut max_depth: i32 = 0;
+ let mut stack = vec![root.unwrap()];
+ while !stack.is_empty() {
+ let num = stack.len();
+ for _i in 0..num{
+ let top = stack.remove(0);
+ if top.borrow_mut().left.is_some(){
+ stack.push(top.borrow_mut().left.take().unwrap());
+ }
+ if top.borrow_mut().right.is_some(){
+ stack.push(top.borrow_mut().right.take().unwrap());
+ }
+ }
+ max_depth+=1;
+ }
+ max_depth
+ }
+```
+
+
那么我们可以顺便解决一下n叉树的最大深度问题
# 559.n叉树的最大深度
diff --git a/problems/0111.二叉树的最小深度.md b/problems/0111.二叉树的最小深度.md
index 224caa5e..b1331659 100644
--- a/problems/0111.二叉树的最小深度.md
+++ b/problems/0111.二叉树的最小深度.md
@@ -488,5 +488,69 @@ func minDepth(_ root: TreeNode?) -> Int {
}
```
+rust:
+```rust
+impl Solution {
+ pub fn min_depth(root: Option>>) -> i32 {
+ return Solution::bfs(root)
+ }
+
+ // 递归
+ pub fn dfs(node: Option>>) -> i32{
+ if node.is_none(){
+ return 0;
+ }
+ let parent = node.unwrap();
+ let left_child = parent.borrow_mut().left.take();
+ let right_child = parent.borrow_mut().right.take();
+ if left_child.is_none() && right_child.is_none(){
+ return 1;
+ }
+ let mut min_depth = i32::MAX;
+ if left_child.is_some(){
+ let left_depth = Solution::dfs(left_child);
+ if left_depth <= min_depth{
+ min_depth = left_depth
+ }
+ }
+ if right_child.is_some(){
+ let right_depth = Solution::dfs(right_child);
+ if right_depth <= min_depth{
+ min_depth = right_depth
+ }
+ }
+ min_depth + 1
+
+ }
+
+ // 迭代
+ pub fn bfs(node: Option>>) -> i32{
+ let mut min_depth = 0;
+ if node.is_none(){
+ return min_depth
+ }
+ let mut stack = vec![node.unwrap()];
+ while !stack.is_empty(){
+ min_depth += 1;
+ let num = stack.len();
+ for _i in 0..num{
+ let top = stack.remove(0);
+ let left_child = top.borrow_mut().left.take();
+ let right_child = top.borrow_mut().right.take();
+ if left_child.is_none() && right_child.is_none(){
+ return min_depth;
+ }
+ if left_child.is_some(){
+ stack.push(left_child.unwrap());
+ }
+ if right_child.is_some(){
+ stack.push(right_child.unwrap());
+ }
+ }
+ }
+ min_depth
+ }
+```
+
-----------------------
diff --git a/problems/0344.反转字符串.md b/problems/0344.反转字符串.md
index 58bada05..e1f27bd7 100644
--- a/problems/0344.反转字符串.md
+++ b/problems/0344.反转字符串.md
@@ -266,6 +266,20 @@ public class Solution
}
}
```
-
+Scala:
+```scala
+object Solution {
+ def reverseString(s: Array[Char]): Unit = {
+ var (left, right) = (0, s.length - 1)
+ while (left < right) {
+ var tmp = s(left)
+ s(left) = s(right)
+ s(right) = tmp
+ left += 1
+ right -= 1
+ }
+ }
+}
+```
-----------------------
diff --git a/problems/0383.赎金信.md b/problems/0383.赎金信.md
index 933a7c31..75dafb72 100644
--- a/problems/0383.赎金信.md
+++ b/problems/0383.赎金信.md
@@ -363,6 +363,70 @@ impl Solution {
}
```
+Scala:
+
+版本一: 使用数组作为哈希表
+```scala
+object Solution {
+ def canConstruct(ransomNote: String, magazine: String): Boolean = {
+ // 如果magazine的长度小于ransomNote的长度，必然是false
+ if (magazine.length < ransomNote.length) {
+ return false
+ }
+ // 定义一个数组，存储magazine字符出现的次数
+ val map: Array[Int] = new Array[Int](26)
+ // 遍历magazine字符串，对应的字符+=1
+ for (i <- magazine.indices) {
+ map(magazine(i) - 'a') += 1
+ }
+ // 遍历ransomNote
+ for (i <- ransomNote.indices) {
+ if (map(ransomNote(i) - 'a') > 0)
+ map(ransomNote(i) - 'a') -= 1
+ else return false
+ }
+ // 如果上面没有返回false，直接返回true，关键字return可以省略
+ true
+ }
+}
+```
+
+```scala
+object Solution {
+ import scala.collection.mutable
+ def canConstruct(ransomNote: String, magazine: String): Boolean = {
+ // 如果magazine的长度小于ransomNote的长度，必然是false
+ if (magazine.length < ransomNote.length) {
+ return false
+ }
+ // 定义map，key是字符，value是字符出现的次数
+ val map = new mutable.HashMap[Char, Int]()
+ // 遍历magazine，把所有的字符都记录到map里面
+ for (i <- magazine.indices) {
+ val tmpChar = magazine(i)
+ // 如果map包含该字符，那么对应的value++，否则添加该字符
+ if (map.contains(tmpChar)) {
+ map.put(tmpChar, map.get(tmpChar).get + 1)
+ } else {
+ map.put(tmpChar, 1)
+ }
+ }
+ // 遍历ransomNote
+ for (i <- ransomNote.indices) {
+ val tmpChar = ransomNote(i)
+ // 如果map包含并且该字符的value大于0，则匹配成功，map对应的--，否则直接返回false
+ if (map.contains(tmpChar) && map.get(tmpChar).get > 0) {
+ map.put(tmpChar, map.get(tmpChar).get - 1)
+ } else {
+ return false
+ }
+ }
+ // 如果上面没有返回false，直接返回true，关键字return可以省略
+ true
+ }
+}
+
+
C#：
```csharp
public bool CanConstruct(string ransomNote, string magazine) {
@@ -379,6 +443,7 @@ public bool CanConstruct(string ransomNote, string magazine) {
}
return true;
}
+
```
-----------------------
diff --git a/problems/0454.四数相加II.md b/problems/0454.四数相加II.md
index 962fe7a5..bfdee26e 100644
--- a/problems/0454.四数相加II.md
+++ b/problems/0454.四数相加II.md
@@ -318,6 +318,43 @@ impl Solution {
}
```
+
+Scala:
+```scala
+object Solution {
+ // 导包
+ import scala.collection.mutable
+ def fourSumCount(nums1: Array[Int], nums2: Array[Int], nums3: Array[Int], nums4: Array[Int]): Int = {
+ // 定义一个HashMap，key存储值，value存储该值出现的次数
+ val map = new mutable.HashMap[Int, Int]()
+ // 遍历前两个数组，把他们所有可能的情况都记录到map
+ for (i <- nums1.indices) {
+ for (j <- nums2.indices) {
+ val tmp = nums1(i) + nums2(j)
+ // 如果包含该值，则对他的key加1，不包含则添加进去
+ if (map.contains(tmp)) {
+ map.put(tmp, map.get(tmp).get + 1)
+ } else {
+ map.put(tmp, 1)
+ }
+ }
+ }
+ var res = 0 // 结果变量
+ // 遍历后两个数组
+ for (i <- nums3.indices) {
+ for (j <- nums4.indices) {
+ val tmp = -(nums3(i) + nums4(j))
+ // 如果map中存在该值，结果就+=value
+ if (map.contains(tmp)) {
+ res += map.get(tmp).get
+ }
+ }
+ }
+ // 返回最终结果，可以省略关键字return
+ res
+ }
+}
+
C#：
```csharp
public int FourSumCount(int[] nums1, int[] nums2, int[] nums3, int[] nums4) {
diff --git a/problems/0541.反转字符串II.md b/problems/0541.反转字符串II.md
index 8c13a390..99d6ebe0 100644
--- a/problems/0541.反转字符串II.md
+++ b/problems/0541.反转字符串II.md
@@ -347,6 +347,47 @@ public class Solution
}
}
```
+Scala:
+版本一: (正常解法)
+```scala
+object Solution {
+ def reverseStr(s: String, k: Int): String = {
+ val res = s.toCharArray // 转换为Array好处理
+ for (i <- s.indices by 2 * k) {
+ // 如果i+k大于了res的长度，则需要全部翻转
+ if (i + k > res.length) {
+ reverse(res, i, s.length - 1)
+ } else {
+ reverse(res, i, i + k - 1)
+ }
+ }
+ new String(res)
+ }
+ // 翻转字符串，从start到end
+ def reverse(s: Array[Char], start: Int, end: Int): Unit = {
+ var (left, right) = (start, end)
+ while (left < right) {
+ var tmp = s(left)
+ s(left) = s(right)
+ s(right) = tmp
+ left += 1
+ right -= 1
+ }
+ }
+}
+```
+版本二: 首先利用sliding每隔k个进行分割，随后转换为数组，再使用zipWithIndex添加每个数组的索引，紧接着利用map做变换，如果索引%2==0则说明需要翻转，否则原封不动，最后再转换为String
+```scala
+object Solution {
+ def reverseStr(s: String, k: Int): String = {
+ s.sliding(k, k)
+ .toArray
+ .zipWithIndex
+ .map(v => if (v._2 % 2 == 0) v._1.reverse else v._1)
+ .mkString
+ }
+}
+```
-----------------------
diff --git a/problems/0714.买卖股票的最佳时机含手续费（动态规划）.md b/problems/0714.买卖股票的最佳时机含手续费（动态规划）.md
index 4ab63e79..5625604b 100644
--- a/problems/0714.买卖股票的最佳时机含手续费（动态规划）.md
+++ b/problems/0714.买卖股票的最佳时机含手续费（动态规划）.md
@@ -200,6 +200,29 @@ const maxProfit = (prices,fee) => {
}
```
+TypeScript：
+
+```typescript
+function maxProfit(prices: number[], fee: number): number {
+ /**
+ dp[i][0]：持有股票
+ dp[i][1]: 不持有
+ */
+ const length: number = prices.length;
+ if (length === 0) return 0;
+ const dp: number[][] = new Array(length).fill(0).map(_ => []);
+ dp[0][0] = -prices[0];
+ dp[0][1] = 0;
+ for (let i = 1; i < length; i++) {
+ dp[i][0] = Math.max(dp[i - 1][0], dp[i - 1][1] - prices[i]);
+ dp[i][1] = Math.max(dp[i - 1][1], dp[i - 1][0] + prices[i] - fee);
+ }
+ return dp[length - 1][1];
+};
+```
+
+
+
-----------------------
diff --git a/problems/剑指Offer05.替换空格.md b/problems/剑指Offer05.替换空格.md
index 037bd427..eecd7f0c 100644
--- a/problems/剑指Offer05.替换空格.md
+++ b/problems/剑指Offer05.替换空格.md
@@ -413,8 +413,62 @@ func replaceSpace(_ s: String) -> String {
}
```
+Scala:
-
+方式一: 双指针
+```scala
+object Solution {
+ def replaceSpace(s: String): String = {
+ var count = 0
+ s.foreach(c => if (c == ' ') count += 1) // 统计空格的数量
+ val sOldSize = s.length // 旧数组字符串长度
+ val sNewSize = s.length + count * 2 // 新数组字符串长度
+ val res = new Array[Char](sNewSize) // 新数组
+ var index = sNewSize - 1 // 新数组索引
+ // 逆序遍历
+ for (i <- (0 until sOldSize).reverse) {
+ if (s(i) == ' ') {
+ res(index) = '0'
+ index -= 1
+ res(index) = '2'
+ index -= 1
+ res(index) = '%'
+ } else {
+ res(index) = s(i)
+ }
+ index -= 1
+ }
+ res.mkString
+ }
+}
+```
+方式二: 使用一个集合，遇到空格就添加%20
+```scala
+object Solution {
+ import scala.collection.mutable.ListBuffer
+ def replaceSpace(s: String): String = {
+ val res: ListBuffer[Char] = ListBuffer[Char]()
+ for (i <- s.indices) {
+ if (s(i) == ' ') {
+ res += '%'
+ res += '2'
+ res += '0'
+ }else{
+ res += s(i)
+ }
+ }
+ res.mkString
+ }
+}
+```
+方式三: 使用map
+```scala
+object Solution {
+ def replaceSpace(s: String): String = {
+ s.map(c => if(c == ' ') "%20" else c).mkString
+ }
+}
+```
-----------------------