diff --git a/problems/0027.移除元素.md b/problems/0027.移除元素.md index 03c58b43..9d687cfb 100644 --- a/problems/0027.移除元素.md +++ b/problems/0027.移除元素.md @@ -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; + } +} +``` + -----------------------
diff --git a/problems/0034.在排序数组中查找元素的第一个和最后一个位置.md b/problems/0034.在排序数组中查找元素的第一个和最后一个位置.md index 260462c2..a81b3641 100644 --- a/problems/0034.在排序数组中查找元素的第一个和最后一个位置.md +++ b/problems/0034.在排序数组中查找元素的第一个和最后一个位置.md @@ -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 { } ``` + -----------------------
diff --git a/problems/0077.组合.md b/problems/0077.组合.md index 8d22d018..fc72be15 100644 --- a/problems/0077.组合.md +++ b/problems/0077.组合.md @@ -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 + } +} +``` + -----------------------
diff --git a/problems/0077.组合优化.md b/problems/0077.组合优化.md index a6767047..8d742566 100644 --- a/problems/0077.组合优化.md +++ b/problems/0077.组合优化.md @@ -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 + } +} +``` + -----------------------
diff --git a/problems/0108.将有序数组转换为二叉搜索树.md b/problems/0108.将有序数组转换为二叉搜索树.md index c9c1a693..b5f322f0 100644 --- a/problems/0108.将有序数组转换为二叉搜索树.md +++ b/problems/0108.将有序数组转换为二叉搜索树.md @@ -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) + } +} +``` + + -----------------------
diff --git a/problems/0150.逆波兰表达式求值.md b/problems/0150.逆波兰表达式求值.md index 05916135..1a90265a 100644 --- a/problems/0150.逆波兰表达式求值.md +++ b/problems/0150.逆波兰表达式求值.md @@ -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;$ipush($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() } + } ``` ----------------------- diff --git a/problems/0235.二叉搜索树的最近公共祖先.md b/problems/0235.二叉搜索树的最近公共祖先.md index 9ff7e293..ee86d02f 100644 --- a/problems/0235.二叉搜索树的最近公共祖先.md +++ b/problems/0235.二叉搜索树的最近公共祖先.md @@ -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 + } +} +``` ----------------------- diff --git a/problems/0236.二叉树的最近公共祖先.md b/problems/0236.二叉树的最近公共祖先.md index 23695b11..c3e2ae7a 100644 --- a/problems/0236.二叉树的最近公共祖先.md +++ b/problems/0236.二叉树的最近公共祖先.md @@ -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 + } +} +``` -----------------------
diff --git a/problems/0239.滑动窗口最大值.md b/problems/0239.滑动窗口最大值.md index 23e79c80..7ee1fdb1 100644 --- a/problems/0239.滑动窗口最大值.md +++ b/problems/0239.滑动窗口最大值.md @@ -654,8 +654,7 @@ object Solution { // 最终返回res,return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"; + } +} +``` + -----------------------
diff --git a/problems/0450.删除二叉搜索树中的节点.md b/problems/0450.删除二叉搜索树中的节点.md index aca9084f..3fa2a1c5 100644 --- a/problems/0450.删除二叉搜索树中的节点.md +++ b/problems/0450.删除二叉搜索树中的节点.md @@ -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 + } +} +``` -----------------------
diff --git a/problems/0538.把二叉搜索树转换为累加树.md b/problems/0538.把二叉搜索树转换为累加树.md index 853cca6f..5c1e9e8c 100644 --- a/problems/0538.把二叉搜索树转换为累加树.md +++ b/problems/0538.把二叉搜索树转换为累加树.md @@ -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 + } +} +``` ----------------------- diff --git a/problems/0669.修剪二叉搜索树.md b/problems/0669.修剪二叉搜索树.md index 154ba5a9..a286315d 100644 --- a/problems/0669.修剪二叉搜索树.md +++ b/problems/0669.修剪二叉搜索树.md @@ -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 + } +} +``` ----------------------- diff --git a/problems/0701.二叉搜索树中的插入操作.md b/problems/0701.二叉搜索树中的插入操作.md index 102f091e..06e1c88f 100644 --- a/problems/0701.二叉搜索树中的插入操作.md +++ b/problems/0701.二叉搜索树中的插入操作.md @@ -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 + } +} +``` + -----------------------
diff --git a/problems/0704.二分查找.md b/problems/0704.二分查找.md index 6a37e4d1..f3b9326d 100644 --- a/problems/0704.二分查找.md +++ b/problems/0704.二分查找.md @@ -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:** diff --git a/problems/0922.按奇偶排序数组II.md b/problems/0922.按奇偶排序数组II.md index 8ca46db9..49547a15 100644 --- a/problems/0922.按奇偶排序数组II.md +++ b/problems/0922.按奇偶排序数组II.md @@ -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; +}; +``` + + + -----------------------
diff --git a/problems/0977.有序数组的平方.md b/problems/0977.有序数组的平方.md index 4052c570..458107dd 100644 --- a/problems/0977.有序数组的平方.md +++ b/problems/0977.有序数组的平方.md @@ -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; + } +} +``` -----------------------