[pull] master from youngyangyang04:master #8

Original file line number	Diff line number	Diff line change
Expand Up		@@ -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 {
Expand All		@@ -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>

45 changes: 44 additions & 1 deletion problems/0015.三数之和.md

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Original file line number	Diff line number	Diff line change
Expand Up		@@ -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>

45 changes: 44 additions & 1 deletion problems/0018.四数之和.md

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Original file line number	Diff line number	Diff line change
Expand Up		@@ -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>

59 changes: 59 additions & 0 deletions problems/0102.二叉树的层序遍历.md

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Original file line number	Diff line number	Diff line change
Expand Up		@@ -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!


Expand Down Expand Up		@@ -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/)
Expand Down

27 changes: 27 additions & 0 deletions problems/0104.二叉树的最大深度.md

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Original file line number	Diff line number	Diff line change
Expand Up		@@ -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
Expand Down

64 changes: 64 additions & 0 deletions problems/0111.二叉树的最小深度.md

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Original file line number	Diff line number	Diff line change
Expand Up		@@ -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>

16 changes: 15 additions & 1 deletion problems/0344.反转字符串.md

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Original file line number	Diff line number	Diff line change
Expand Up		@@ -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>

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