mirror of
https://github.com/youngyangyang04/leetcode-master.git
synced 2025-07-09 03:34:02 +08:00
Merge branch 'youngyangyang04:master' into master
This commit is contained in:
@ -557,6 +557,37 @@ func letterCombinations(_ digits: String) -> [String] {
|
||||
}
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||||
```
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## Scala:
|
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|
||||
```scala
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object Solution {
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import scala.collection.mutable
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def letterCombinations(digits: String): List[String] = {
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var result = mutable.ListBuffer[String]()
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if(digits == "") return result.toList // 如果参数为空,返回空结果集的List形式
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var path = mutable.ListBuffer[Char]()
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// 数字和字符的映射关系
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val map = Array[String]("", "", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz")
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def backtracking(index: Int): Unit = {
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if (index == digits.size) {
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result.append(path.mkString) // mkString语法:将数组类型直接转换为字符串
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return
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}
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var digit = digits(index) - '0' // 这里使用toInt会报错!必须 -'0'
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for (i <- 0 until map(digit).size) {
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path.append(map(digit)(i))
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backtracking(index + 1)
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path = path.take(path.size - 1)
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}
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}
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backtracking(0)
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result.toList
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}
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}
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||||
```
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||||
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||||
|
||||
-----------------------
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||||
<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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||||
|
@ -339,7 +339,6 @@ int removeElement(int* nums, int numsSize, int val){
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}
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||||
```
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||||
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||||
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||||
Kotlin:
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||||
```kotlin
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fun removeElement(nums: IntArray, `val`: Int): Int {
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||||
@ -351,7 +350,6 @@ fun removeElement(nums: IntArray, `val`: Int): Int {
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||||
}
|
||||
```
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||||
|
||||
|
||||
Scala:
|
||||
```scala
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||||
object Solution {
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||||
@ -368,5 +366,20 @@ object Solution {
|
||||
}
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||||
```
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||||
|
||||
C#:
|
||||
```csharp
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||||
public class Solution {
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||||
public int RemoveElement(int[] nums, int val) {
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||||
int slow = 0;
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for (int fast = 0; fast < nums.Length; fast++) {
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||||
if (val != nums[fast]) {
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nums[slow++] = nums[fast];
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||||
}
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||||
}
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||||
return slow;
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||||
}
|
||||
}
|
||||
```
|
||||
|
||||
-----------------------
|
||||
<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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||||
|
@ -502,5 +502,35 @@ func combinationSum(_ candidates: [Int], _ target: Int) -> [[Int]] {
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||||
}
|
||||
```
|
||||
|
||||
## Scala
|
||||
|
||||
```scala
|
||||
object Solution {
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||||
import scala.collection.mutable
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||||
def combinationSum(candidates: Array[Int], target: Int): List[List[Int]] = {
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||||
var result = mutable.ListBuffer[List[Int]]()
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||||
var path = mutable.ListBuffer[Int]()
|
||||
|
||||
def backtracking(sum: Int, index: Int): Unit = {
|
||||
if (sum == target) {
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result.append(path.toList) // 如果正好等于target,就添加到结果集
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||||
return
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||||
}
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||||
// 应该是从当前索引开始的,而不是从0
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// 剪枝优化:添加循环守卫,当sum + c(i) <= target的时候才循环,才可以进入下一次递归
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for (i <- index until candidates.size if sum + candidates(i) <= target) {
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path.append(candidates(i))
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backtracking(sum + candidates(i), i)
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||||
path = path.take(path.size - 1)
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||||
}
|
||||
}
|
||||
|
||||
backtracking(0, 0)
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||||
result.toList
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||||
}
|
||||
}
|
||||
```
|
||||
|
||||
|
||||
-----------------------
|
||||
<div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>
|
||||
|
@ -693,5 +693,37 @@ func combinationSum2(_ candidates: [Int], _ target: Int) -> [[Int]] {
|
||||
}
|
||||
```
|
||||
|
||||
|
||||
## Scala
|
||||
|
||||
```scala
|
||||
object Solution {
|
||||
import scala.collection.mutable
|
||||
def combinationSum2(candidates: Array[Int], target: Int): List[List[Int]] = {
|
||||
var res = mutable.ListBuffer[List[Int]]()
|
||||
var path = mutable.ListBuffer[Int]()
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||||
var candidate = candidates.sorted
|
||||
|
||||
def backtracking(sum: Int, startIndex: Int): Unit = {
|
||||
if (sum == target) {
|
||||
res.append(path.toList)
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||||
return
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||||
}
|
||||
|
||||
for (i <- startIndex until candidate.size if sum + candidate(i) <= target) {
|
||||
if (!(i > startIndex && candidate(i) == candidate(i - 1))) {
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||||
path.append(candidate(i))
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||||
backtracking(sum + candidate(i), i + 1)
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||||
path = path.take(path.size - 1)
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||||
}
|
||||
}
|
||||
}
|
||||
|
||||
backtracking(0, 0)
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||||
res.toList
|
||||
}
|
||||
}
|
||||
```
|
||||
|
||||
-----------------------
|
||||
<div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>
|
||||
|
@ -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>
|
||||
|
@ -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>
|
||||
|
@ -502,5 +502,35 @@ func combinationSum3(_ count: Int, _ targetSum: Int) -> [[Int]] {
|
||||
}
|
||||
```
|
||||
|
||||
## Scala
|
||||
|
||||
```scala
|
||||
object Solution {
|
||||
import scala.collection.mutable
|
||||
def combinationSum3(k: Int, n: Int): List[List[Int]] = {
|
||||
var result = mutable.ListBuffer[List[Int]]()
|
||||
var path = mutable.ListBuffer[Int]()
|
||||
|
||||
def backtracking(k: Int, n: Int, sum: Int, startIndex: Int): Unit = {
|
||||
if (sum > n) return // 剪枝,如果sum>目标和,就返回
|
||||
if (sum == n && path.size == k) {
|
||||
result.append(path.toList)
|
||||
return
|
||||
}
|
||||
// 剪枝
|
||||
for (i <- startIndex to (9 - (k - path.size) + 1)) {
|
||||
path.append(i)
|
||||
backtracking(k, n, sum + i, i + 1)
|
||||
path = path.take(path.size - 1)
|
||||
}
|
||||
}
|
||||
|
||||
backtracking(k, n, 0, 1) // 调用递归方法
|
||||
result.toList // 最终返回结果集的List形式
|
||||
}
|
||||
}
|
||||
```
|
||||
|
||||
|
||||
-----------------------
|
||||
<div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>
|
||||
|
@ -654,7 +654,6 @@ object Solution {
|
||||
// 最终返回res,return关键字可以省略
|
||||
res
|
||||
}
|
||||
|
||||
}
|
||||
|
||||
class MyQueue {
|
||||
@ -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>
|
||||
|
@ -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
|
||||
}
|
||||
}
|
||||
```
|
||||
|
||||
|
||||
-----------------------
|
||||
|
@ -104,8 +104,9 @@ public:
|
||||
// 在第index个节点之前插入一个新节点,例如index为0,那么新插入的节点为链表的新头节点。
|
||||
// 如果index 等于链表的长度,则说明是新插入的节点为链表的尾结点
|
||||
// 如果index大于链表的长度,则返回空
|
||||
// 如果index小于0,则置为0,作为链表的新头节点。
|
||||
void addAtIndex(int index, int val) {
|
||||
if (index > _size) {
|
||||
if (index > _size || index < 0) {
|
||||
return;
|
||||
}
|
||||
LinkedNode* newNode = new LinkedNode(val);
|
||||
|
@ -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>
|
||||
|
@ -356,10 +356,14 @@ func test_2_wei_bag_problem1(weight, value []int, bagweight int) int {
|
||||
// 递推公式
|
||||
for i := 1; i < len(weight); i++ {
|
||||
//正序,也可以倒序
|
||||
for j := weight[i];j<= bagweight ; j++ {
|
||||
for j := 0; j <= bagweight; j++ {
|
||||
if j < weight[i] {
|
||||
dp[i][j] = dp[i-1][j]
|
||||
} else {
|
||||
dp[i][j] = max(dp[i-1][j], dp[i-1][j-weight[i]]+value[i])
|
||||
}
|
||||
}
|
||||
}
|
||||
return dp[len(weight)-1][bagweight]
|
||||
}
|
||||
|
||||
|
Reference in New Issue
Block a user