Merge branch 'youngyangyang04:master' into master

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 {
 }
 ```
 
+
 -----------------------
 <div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

problems/0015.三数之和.md

@@ -616,6 +616,49 @@ public class Solution
     }
 }
 ```
+Scala:
+```scala
+object Solution {
+  // 导包
+  import scala.collection.mutable.ListBuffer
+  import scala.util.control.Breaks.{break, breakable}
 
+  def threeSum(nums: Array[Int]): List[List[Int]] = {
+    // 定义结果集，最后需要转换为List
+    val res = ListBuffer[List[Int]]()
+    val nums_tmp = nums.sorted // 对nums进行排序
+    for (i <- nums_tmp.indices) {
+      // 如果要排的第一个数字大于0，直接返回结果
+      if (nums_tmp(i) > 0) {
+        return res.toList
+      }
+      // 如果i大于0并且和前一个数字重复，则跳过本次循环，相当于continue
+      breakable {
+        if (i > 0 && nums_tmp(i) == nums_tmp(i - 1)) {
+          break
+        } else {
+          var left = i + 1
+          var right = nums_tmp.length - 1
+          while (left < right) {
+            var sum = nums_tmp(i) + nums_tmp(left) + nums_tmp(right) // 求三数之和
+            if (sum < 0) left += 1
+            else if (sum > 0) right -= 1
+            else {
+              res += List(nums_tmp(i), nums_tmp(left), nums_tmp(right)) // 如果等于0 添加进结果集
+              // 为了避免重复，对left和right进行移动
+              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
+            }
+          }
+        }
+      }
+    }
+    // 最终返回需要转换为List，return关键字可以省略
+    res.toList
+  }
+}
+```
 -----------------------
 <div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

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
+  }
+}
+```
 -----------------------
 <div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

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<Rc<RefCell<TreeNode>>>) -> Vec<Vec<i32>> {
+    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<Rc<RefCell<TreeNode>>>) -> Vec<Vec<i32>> {
+    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/)

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<Rc<RefCell<TreeNode>>>) -> 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

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<Rc<RefCell<TreeNode>>>) -> i32 {
+        return Solution::bfs(root)
+    }
+
+    // 递归
+    pub fn dfs(node: Option<Rc<RefCell<TreeNode>>>) -> 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<Rc<RefCell<TreeNode>>>) -> 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
+    }
+```
+
 -----------------------
 <div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

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
+    }
+  }
+}
+```
 -----------------------
 <div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

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;
     }
+
 ```
 -----------------------
 <div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

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) {

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
+  }
+}
+```
 -----------------------
 <div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

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];
+};
+```
+
+
+
 
 -----------------------
 <div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

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
+  }
+}
+```
 
 
 -----------------------