Merge branch 'youngyangyang04:master' into master

This commit is contained in:
w2xi
2022-07-06 21:53:52 +08:00
committed by GitHub
16 changed files with 551 additions and 5 deletions

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@ -339,7 +339,6 @@ int removeElement(int* nums, int numsSize, int val){
}
```
Kotlin:
```kotlin
fun removeElement(nums: IntArray, `val`: Int): Int {
@ -351,7 +350,6 @@ fun removeElement(nums: IntArray, `val`: Int): Int {
}
```
Scala:
```scala
object Solution {
@ -368,5 +366,20 @@ object Solution {
}
```
C#:
```csharp
public class Solution {
public int RemoveElement(int[] nums, int val) {
int slow = 0;
for (int fast = 0; fast < nums.Length; fast++) {
if (val != nums[fast]) {
nums[slow++] = nums[fast];
}
}
return slow;
}
}
```
-----------------------
<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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@ -480,6 +480,62 @@ var searchRange = function(nums, target) {
return [-1, -1];
};
```
### TypeScript
```typescript
function searchRange(nums: number[], target: number): number[] {
const leftBoard: number = getLeftBorder(nums, target);
const rightBoard: number = getRightBorder(nums, target);
// target 在nums区间左侧或右侧
if (leftBoard === (nums.length - 1) || rightBoard === 0) return [-1, -1];
// target 不存在与nums范围内
if (rightBoard - leftBoard <= 1) return [-1, -1];
// target 存在于nums范围内
return [leftBoard + 1, rightBoard - 1];
};
// 查找第一个大于target的元素下标
function getRightBorder(nums: number[], target: number): number {
let left: number = 0,
right: number = nums.length - 1;
// 0表示target在nums区间的左边
let rightBoard: number = 0;
while (left <= right) {
let mid = Math.floor((left + right) / 2);
if (nums[mid] <= target) {
// 右边界一定在mid右边不含mid
left = mid + 1;
rightBoard = left;
} else {
// 右边界在mid左边含mid
right = mid - 1;
}
}
return rightBoard;
}
// 查找第一个小于target的元素下标
function getLeftBorder(nums: number[], target: number): number {
let left: number = 0,
right: number = nums.length - 1;
// length-1表示target在nums区间的右边
let leftBoard: number = nums.length - 1;
while (left <= right) {
let mid = Math.floor((left + right) / 2);
if (nums[mid] >= target) {
// 左边界一定在mid左边不含mid
right = mid - 1;
leftBoard = right;
} else {
// 左边界在mid右边含mid
left = mid + 1;
}
}
return leftBoard;
}
```
### Scala
```scala
object Solution {
@ -527,5 +583,6 @@ object Solution {
}
```
-----------------------
<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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@ -673,5 +673,63 @@ func combine(_ n: Int, _ k: Int) -> [[Int]] {
}
```
### Scala
暴力:
```scala
object Solution {
import scala.collection.mutable // 导包
def combine(n: Int, k: Int): List[List[Int]] = {
var result = mutable.ListBuffer[List[Int]]() // 存放结果集
var path = mutable.ListBuffer[Int]() //存放符合条件的结果
def backtracking(n: Int, k: Int, startIndex: Int): Unit = {
if (path.size == k) {
// 如果path的size == k就达到题目要求添加到结果集并返回
result.append(path.toList)
return
}
for (i <- startIndex to n) { // 遍历从startIndex到n
path.append(i) // 先把数字添加进去
backtracking(n, k, i + 1) // 进行下一步回溯
path = path.take(path.size - 1) // 回溯完再删除掉刚刚添加的数字
}
}
backtracking(n, k, 1) // 执行回溯
result.toList // 最终返回result的List形式return关键字可以省略
}
}
```
剪枝:
```scala
object Solution {
import scala.collection.mutable // 导包
def combine(n: Int, k: Int): List[List[Int]] = {
var result = mutable.ListBuffer[List[Int]]() // 存放结果集
var path = mutable.ListBuffer[Int]() //存放符合条件的结果
def backtracking(n: Int, k: Int, startIndex: Int): Unit = {
if (path.size == k) {
// 如果path的size == k就达到题目要求添加到结果集并返回
result.append(path.toList)
return
}
// 剪枝优化
for (i <- startIndex to (n - (k - path.size) + 1)) {
path.append(i) // 先把数字添加进去
backtracking(n, k, i + 1) // 进行下一步回溯
path = path.take(path.size - 1) // 回溯完再删除掉刚刚添加的数字
}
}
backtracking(n, k, 1) // 执行回溯
result.toList // 最终返回result的List形式return关键字可以省略
}
}
```
-----------------------
<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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@ -346,5 +346,34 @@ func combine(_ n: Int, _ k: Int) -> [[Int]] {
}
```
Scala:
```scala
object Solution {
import scala.collection.mutable // 导包
def combine(n: Int, k: Int): List[List[Int]] = {
var result = mutable.ListBuffer[List[Int]]() // 存放结果集
var path = mutable.ListBuffer[Int]() //存放符合条件的结果
def backtracking(n: Int, k: Int, startIndex: Int): Unit = {
if (path.size == k) {
// 如果path的size == k就达到题目要求添加到结果集并返回
result.append(path.toList)
return
}
// 剪枝优化
for (i <- startIndex to (n - (k - path.size) + 1)) {
path.append(i) // 先把数字添加进去
backtracking(n, k, i + 1) // 进行下一步回溯
path = path.take(path.size - 1) // 回溯完再删除掉刚刚添加的数字
}
}
backtracking(n, k, 1) // 执行回溯
result.toList // 最终返回result的List形式return关键字可以省略
}
}
```
-----------------------
<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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@ -448,5 +448,27 @@ struct TreeNode* sortedArrayToBST(int* nums, int numsSize) {
}
```
## Scala
递归:
```scala
object Solution {
def sortedArrayToBST(nums: Array[Int]): TreeNode = {
def buildTree(left: Int, right: Int): TreeNode = {
if (left > right) return null // 当left大于right的时候返回空
// 最中间的节点是当前节点
var mid = left + (right - left) / 2
var curNode = new TreeNode(nums(mid))
curNode.left = buildTree(left, mid - 1)
curNode.right = buildTree(mid + 1, right)
curNode
}
buildTree(0, nums.size - 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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@ -325,6 +325,33 @@ func evalRPN(_ tokens: [String]) -> Int {
return stack.last!
}
```
PHP
```php
class Solution {
function evalRPN($tokens) {
$st = new SplStack();
for($i = 0;$i<count($tokens);$i++){
// 是数字直接入栈
if(is_numeric($tokens[$i])){
$st->push($tokens[$i]);
}else{
// 是符号进行运算
$num1 = $st->pop();
$num2 = $st->pop();
if ($tokens[$i] == "+") $st->push($num2 + $num1);
if ($tokens[$i] == "-") $st->push($num2 - $num1);
if ($tokens[$i] == "*") $st->push($num2 * $num1);
// 注意处理小数部分
if ($tokens[$i] == "/") $st->push(intval($num2 / $num1));
}
}
return $st->pop();
}
}
```
Scala:
```scala
object Solution {
@ -351,6 +378,7 @@ object Solution {
// 最后返回栈顶不需要加return关键字
stack.pop()
}
}
```
-----------------------

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@ -381,7 +381,36 @@ function lowestCommonAncestor(root: TreeNode | null, p: TreeNode | null, q: Tree
};
```
## Scala
递归:
```scala
object Solution {
def lowestCommonAncestor(root: TreeNode, p: TreeNode, q: TreeNode): TreeNode = {
// scala中每个关键字都有其返回值于是可以不写return
if (root.value > p.value && root.value > q.value) lowestCommonAncestor(root.left, p, q)
else if (root.value < p.value && root.value < q.value) lowestCommonAncestor(root.right, p, q)
else root
}
}
```
迭代:
```scala
object Solution {
def lowestCommonAncestor(root: TreeNode, p: TreeNode, q: TreeNode): TreeNode = {
var curNode = root // 因为root是不可变量所以要赋值给curNode一个可变量
while(curNode != null){
if(curNode.value > p.value && curNode.value > q.value) curNode = curNode.left
else if(curNode.value < p.value && curNode.value < q.value) curNode = curNode.right
else return curNode
}
null
}
}
```
-----------------------

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@ -343,7 +343,25 @@ function lowestCommonAncestor(root: TreeNode | null, p: TreeNode | null, q: Tree
};
```
## Scala
```scala
object Solution {
def lowestCommonAncestor(root: TreeNode, p: TreeNode, q: TreeNode): TreeNode = {
// 递归结束条件
if (root == null || root == p || root == q) {
return root
}
var left = lowestCommonAncestor(root.left, p, q)
var right = lowestCommonAncestor(root.right, p, q)
if (left != null && right != null) return root
if (left == null) return right
left
}
}
```
-----------------------
<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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@ -654,8 +654,7 @@ object Solution {
// 最终返回resreturn关键字可以省略
res
}
}
}
class MyQueue {
var queue = ArrayBuffer[Int]()
@ -678,5 +677,84 @@ class MyQueue {
def peek(): Int = queue.head
}
```
PHP:
```php
class Solution {
/**
* @param Integer[] $nums
* @param Integer $k
* @return Integer[]
*/
function maxSlidingWindow($nums, $k) {
$myQueue = new MyQueue();
// 先将前k的元素放进队列
for ($i = 0; $i < $k; $i++) {
$myQueue->push($nums[$i]);
}
$result = [];
$result[] = $myQueue->max(); // result 记录前k的元素的最大值
for ($i = $k; $i < count($nums); $i++) {
$myQueue->pop($nums[$i - $k]); // 滑动窗口移除最前面元素
$myQueue->push($nums[$i]); // 滑动窗口前加入最后面的元素
$result[]= $myQueue->max(); // 记录对应的最大值
}
return $result;
}
}
// 单调对列构建
class MyQueue{
private $queue;
public function __construct(){
$this->queue = new SplQueue(); //底层是双向链表实现。
}
public function pop($v){
// 判断当前对列是否为空
// 比较当前要弹出的数值是否等于队列出口元素的数值,如果相等则弹出。
// bottom 从链表前端查看元素, dequeue 从双向链表的开头移动一个节点
if(!$this->queue->isEmpty() && $v == $this->queue->bottom()){
$this->queue->dequeue(); //弹出队列
}
}
public function push($v){
// 判断当前对列是否为空
// 如果push的数值大于入口元素的数值那么就将队列后端的数值弹出直到push的数值小于等于队列入口元素的数值为止。
// 这样就保持了队列里的数值是单调从大到小的了。
while (!$this->queue->isEmpty() && $v > $this->queue->top()) {
$this->queue->pop(); // pop从链表末尾弹出一个元素
}
$this->queue->enqueue($v);
}
// 查询当前队列里的最大值 直接返回队首
public function max(){
// bottom 从链表前端查看元素, top从链表末尾查看元素
return $this->queue->bottom();
}
// 辅助理解: 打印队列元素
public function println(){
// "迭代器移动到链表头部": 可理解为从头遍历链表元素做准备。
// 【PHP中没有指针概念所以就没说指针。从数据结构上理解就是把指针指向链表头部】
$this->queue->rewind();
echo "Println: ";
while($this->queue->valid()){
echo $this->queue->current()," -> ";
$this->queue->next();
}
echo "\n";
}
}
```
-----------------------
<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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@ -582,7 +582,35 @@ function deleteNode(root: TreeNode | null, key: number): TreeNode | null {
};
```
## Scala
```scala
object Solution {
def deleteNode(root: TreeNode, key: Int): TreeNode = {
if (root == null) return root // 第一种情况,没找到删除的节点,遍历到空节点直接返回
if (root.value == key) {
// 第二种情况: 左右孩子都为空直接删除节点返回null
if (root.left == null && root.right == null) return null
// 第三种情况: 左孩子为空,右孩子不为空,右孩子补位
else if (root.left == null && root.right != null) return root.right
// 第四种情况: 左孩子不为空,右孩子为空,左孩子补位
else if (root.left != null && root.right == null) return root.left
// 第五种情况: 左右孩子都不为空,将删除节点的左子树头节点(左孩子)放到
// 右子树的最左边节点的左孩子上,返回删除节点的右孩子
else {
var tmp = root.right
while (tmp.left != null) tmp = tmp.left
tmp.left = root.left
return root.right
}
}
if (root.value > key) root.left = deleteNode(root.left, key)
if (root.value < key) root.right = deleteNode(root.right, key)
root // 返回根节点return关键字可以省略
}
}
```
-----------------------
<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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@ -352,6 +352,24 @@ function convertBST(root: TreeNode | null): TreeNode | null {
};
```
## Scala
```scala
object Solution {
def convertBST(root: TreeNode): TreeNode = {
var sum = 0
def convert(node: TreeNode): Unit = {
if (node == null) return
convert(node.right)
sum += node.value
node.value = sum
convert(node.left)
}
convert(root)
root
}
}
```
-----------------------

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@ -453,7 +453,21 @@ function trimBST(root: TreeNode | null, low: number, high: number): TreeNode | n
};
```
## Scala
递归法:
```scala
object Solution {
def trimBST(root: TreeNode, low: Int, high: Int): TreeNode = {
if (root == null) return null
if (root.value < low) return trimBST(root.right, low, high)
if (root.value > high) return trimBST(root.left, low, high)
root.left = trimBST(root.left, low, high)
root.right = trimBST(root.right, low, high)
root
}
}
```
-----------------------

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@ -585,6 +585,43 @@ function insertIntoBST(root: TreeNode | null, val: number): TreeNode | null {
```
## Scala
递归:
```scala
object Solution {
def insertIntoBST(root: TreeNode, `val`: Int): TreeNode = {
if (root == null) return new TreeNode(`val`)
if (`val` < root.value) root.left = insertIntoBST(root.left, `val`)
else root.right = insertIntoBST(root.right, `val`)
root // 返回根节点
}
}
```
迭代:
```scala
object Solution {
def insertIntoBST(root: TreeNode, `val`: Int): TreeNode = {
if (root == null) {
return new TreeNode(`val`)
}
var parent = root // 记录当前节点的父节点
var curNode = root
while (curNode != null) {
parent = curNode
if(`val` < curNode.value) curNode = curNode.left
else curNode = curNode.right
}
if(`val` < parent.value) parent.left = new TreeNode(`val`)
else parent.right = new TreeNode(`val`)
root // 最终返回根节点
}
}
```
-----------------------
<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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@ -613,6 +613,36 @@ public class Solution{
}
```
**Kotlin:**
```kotlin
class Solution {
fun search(nums: IntArray, target: Int): Int {
// leftBorder
var left:Int = 0
// rightBorder
var right:Int = nums.size - 1
// 使用左闭右闭区间
while (left <= right) {
var middle:Int = left + (right - left)/2
// taget 在左边
if (nums[middle] > target) {
right = middle - 1
}
else {
// target 在右边
if (nums[middle] < target) {
left = middle + 1
}
// 找到了,返回
else return middle
}
}
// 没找到,返回
return -1
}
}
```
**Kotlin:**

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@ -260,6 +260,75 @@ var sortArrayByParityII = function(nums) {
};
```
### TypeScript
> 方法一:
```typescript
function sortArrayByParityII(nums: number[]): number[] {
const evenArr: number[] = [],
oddArr: number[] = [];
for (let num of nums) {
if (num % 2 === 0) {
evenArr.push(num);
} else {
oddArr.push(num);
}
}
const resArr: number[] = [];
for (let i = 0, length = nums.length / 2; i < length; i++) {
resArr.push(evenArr[i]);
resArr.push(oddArr[i]);
}
return resArr;
};
```
> 方法二:
```typescript
function sortArrayByParityII(nums: number[]): number[] {
const length: number = nums.length;
const resArr: number[] = [];
let evenIndex: number = 0,
oddIndex: number = 1;
for (let i = 0; i < length; i++) {
if (nums[i] % 2 === 0) {
resArr[evenIndex] = nums[i];
evenIndex += 2;
} else {
resArr[oddIndex] = nums[i];
oddIndex += 2;
}
}
return resArr;
};
```
> 方法三:
```typescript
function sortArrayByParityII(nums: number[]): number[] {
const length: number = nums.length;
let oddIndex: number = 1;
for (let evenIndex = 0; evenIndex < length; evenIndex += 2) {
if (nums[evenIndex] % 2 === 1) {
// 在偶数位遇到了奇数
while (oddIndex < length && nums[oddIndex] % 2 === 1) {
oddIndex += 2;
}
// 在奇数位遇到了偶数,交换
let temp = nums[evenIndex];
nums[evenIndex] = nums[oddIndex];
nums[oddIndex] = temp;
}
}
return nums;
};
```
-----------------------
<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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@ -420,6 +420,24 @@ object Solution {
}
```
C#
```csharp
public class Solution {
public int[] SortedSquares(int[] nums) {
int k = nums.Length - 1;
int[] result = new int[nums.Length];
for (int i = 0, j = nums.Length - 1;i <= j;){
if (nums[i] * nums[i] < nums[j] * nums[j]) {
result[k--] = nums[j] * nums[j];
j--;
} else {
result[k--] = nums[i] * nums[i];
i++;
}
}
return result;
}
}
```
-----------------------
<div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>