diff --git a/problems/0017.电话号码的字母组合.md b/problems/0017.电话号码的字母组合.md
index 2b5cb978..e357a33b 100644
--- a/problems/0017.电话号码的字母组合.md
+++ b/problems/0017.电话号码的字母组合.md
@@ -557,6 +557,37 @@ func letterCombinations(_ digits: String) -> [String] {
}
```
+## Scala:
+
+```scala
+object Solution {
+ import scala.collection.mutable
+ def letterCombinations(digits: String): List[String] = {
+ var result = mutable.ListBuffer[String]()
+ if(digits == "") return result.toList // 如果参数为空，返回空结果集的List形式
+ var path = mutable.ListBuffer[Char]()
+ // 数字和字符的映射关系
+ val map = Array[String]("", "", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz")
+
+ def backtracking(index: Int): Unit = {
+ if (index == digits.size) {
+ result.append(path.mkString) // mkString语法：将数组类型直接转换为字符串
+ return
+ }
+ var digit = digits(index) - '0' // 这里使用toInt会报错！必须 -'0'
+ for (i <- 0 until map(digit).size) {
+ path.append(map(digit)(i))
+ backtracking(index + 1)
+ path = path.take(path.size - 1)
+ }
+ }
+
+ backtracking(0)
+ result.toList
+ }
+}
+```
+
-----------------------
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/0035.搜索插入位置.md b/problems/0035.搜索插入位置.md
index ed7c49b4..77def4f8 100644
--- a/problems/0035.搜索插入位置.md
+++ b/problems/0035.搜索插入位置.md
@@ -283,6 +283,28 @@ var searchInsert = function (nums, target) {
};
```
+### TypeScript
+
+```typescript
+// 第一种二分法
+function searchInsert(nums: number[], target: number): number {
+ const length: number = nums.length;
+ let left: number = 0,
+ right: number = length - 1;
+ while (left <= right) {
+ const mid: number = Math.floor((left + right) / 2);
+ if (nums[mid] < target) {
+ left = mid + 1;
+ } else if (nums[mid] === target) {
+ return mid;
+ } else {
+ right = mid - 1;
+ }
+ }
+ return right + 1;
+};
+```
+
### Swift
```swift
diff --git a/problems/0039.组合总和.md b/problems/0039.组合总和.md
index fef9b676..564d13ea 100644
--- a/problems/0039.组合总和.md
+++ b/problems/0039.组合总和.md
@@ -502,5 +502,35 @@ func combinationSum(_ candidates: [Int], _ target: Int) -> [[Int]] {
}
```
+## Scala
+
+```scala
+object Solution {
+ import scala.collection.mutable
+ def combinationSum(candidates: Array[Int], target: Int): List[List[Int]] = {
+ var result = mutable.ListBuffer[List[Int]]()
+ var path = mutable.ListBuffer[Int]()
+
+ def backtracking(sum: Int, index: Int): Unit = {
+ if (sum == target) {
+ result.append(path.toList) // 如果正好等于target，就添加到结果集
+ return
+ }
+ // 应该是从当前索引开始的，而不是从0
+ // 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
+ for (i <- index until candidates.size if sum + candidates(i) <= target) {
+ path.append(candidates(i))
+ backtracking(sum + candidates(i), i)
+ path = path.take(path.size - 1)
+ }
+ }
+
+ backtracking(0, 0)
+ result.toList
+ }
+}
+```
+
+
-----------------------
diff --git a/problems/0040.组合总和II.md b/problems/0040.组合总和II.md
index ee4371cd..557eb855 100644
--- a/problems/0040.组合总和II.md
+++ b/problems/0040.组合总和II.md
@@ -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]()
+ var candidate = candidates.sorted
+
+ def backtracking(sum: Int, startIndex: Int): Unit = {
+ if (sum == target) {
+ res.append(path.toList)
+ return
+ }
+
+ for (i <- startIndex until candidate.size if sum + candidate(i) <= target) {
+ if (!(i > startIndex && candidate(i) == candidate(i - 1))) {
+ path.append(candidate(i))
+ backtracking(sum + candidate(i), i + 1)
+ path = path.take(path.size - 1)
+ }
+ }
+ }
+
+ backtracking(0, 0)
+ res.toList
+ }
+}
+```
+
-----------------------
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/0093.复原IP地址.md b/problems/0093.复原IP地址.md
index 41940487..11ca2d03 100644
--- a/problems/0093.复原IP地址.md
+++ b/problems/0093.复原IP地址.md
@@ -659,6 +659,48 @@ func restoreIpAddresses(_ s: String) -> [String] {
}
```
+## Scala
+
+```scala
+object Solution {
+ import scala.collection.mutable
+ def restoreIpAddresses(s: String): List[String] = {
+ var result = mutable.ListBuffer[String]()
+ if (s.size < 4 || s.length > 12) return result.toList
+ var path = mutable.ListBuffer[String]()
+
+ // 判断IP中的一个字段是否为正确的
+ def isIP(sub: String): Boolean = {
+ if (sub.size > 1 && sub(0) == '0') return false
+ if (sub.toInt > 255) return false
+ true
+ }
+
+ def backtracking(startIndex: Int): Unit = {
+ if (startIndex >= s.size) {
+ if (path.size == 4) {
+ result.append(path.mkString(".")) // mkString方法可以把集合里的数据以指定字符串拼接
+ return
+ }
+ return
+ }
+ // subString
+ for (i <- startIndex until startIndex + 3 if i < s.size) {
+ var subString = s.substring(startIndex, i + 1)
+ if (isIP(subString)) { // 如果合法则进行下一轮
+ path.append(subString)
+ backtracking(i + 1)
+ path = path.take(path.size - 1)
+ }
+ }
+ }
+
+ backtracking(0)
+ result.toList
+ }
+}
+```
+
-----------------------
diff --git a/problems/0106.从中序与后序遍历序列构造二叉树.md b/problems/0106.从中序与后序遍历序列构造二叉树.md
index 6d87c756..91e0c8d8 100644
--- a/problems/0106.从中序与后序遍历序列构造二叉树.md
+++ b/problems/0106.从中序与后序遍历序列构造二叉树.md
@@ -584,35 +584,29 @@ tree2 的前序遍历是[1 2 3]， 后序遍历是[3 2 1]。
```java
class Solution {
+ Map map; // 方便根据数值查找位置
public TreeNode buildTree(int[] inorder, int[] postorder) {
- return buildTree1(inorder, 0, inorder.length, postorder, 0, postorder.length);
+ map = new HashMap<>();
+ for (int i = 0; i < inorder.length; i++) { // 用map保存中序序列的数值对应位置
+ map.put(inorder[i], i);
+ }
+
+ return findNode(inorder, 0, inorder.length, postorder,0, postorder.length); // 前闭后开
}
- public TreeNode buildTree1(int[] inorder, int inLeft, int inRight,
- int[] postorder, int postLeft, int postRight) {
- // 没有元素了
- if (inRight - inLeft < 1) {
+
+ public TreeNode findNode(int[] inorder, int inBegin, int inEnd, int[] postorder, int postBegin, int postEnd) {
+ // 参数里的范围都是前闭后开
+ if (inBegin >= inEnd || postBegin >= postEnd) { // 不满足左闭右开，说明没有元素，返回空树
return null;
}
- // 只有一个元素了
- if (inRight - inLeft == 1) {
- return new TreeNode(inorder[inLeft]);
- }
- // 后序数组postorder里最后一个即为根结点
- int rootVal = postorder[postRight - 1];
- TreeNode root = new TreeNode(rootVal);
- int rootIndex = 0;
- // 根据根结点的值找到该值在中序数组inorder里的位置
- for (int i = inLeft; i < inRight; i++) {
- if (inorder[i] == rootVal) {
- rootIndex = i;
- break;
- }
- }
- // 根据rootIndex划分左右子树
- root.left = buildTree1(inorder, inLeft, rootIndex,
- postorder, postLeft, postLeft + (rootIndex - inLeft));
- root.right = buildTree1(inorder, rootIndex + 1, inRight,
- postorder, postLeft + (rootIndex - inLeft), postRight - 1);
+ int rootIndex = map.get(postorder[postEnd - 1]); // 找到后序遍历的最后一个元素在中序遍历中的位置
+ TreeNode root = new TreeNode(inorder[rootIndex]); // 构造结点
+ int lenOfLeft = rootIndex - inBegin; // 保存中序左子树个数，用来确定后序数列的个数
+ root.left = findNode(inorder, inBegin, rootIndex,
+ postorder, postBegin, postBegin + lenOfLeft);
+ root.right = findNode(inorder, rootIndex + 1, inEnd,
+ postorder, postBegin + lenOfLeft, postEnd - 1);
+
return root;
}
}
@@ -622,31 +616,29 @@ class Solution {
```java
class Solution {
+ Map map;
public TreeNode buildTree(int[] preorder, int[] inorder) {
- return helper(preorder, 0, preorder.length - 1, inorder, 0, inorder.length - 1);
- }
-
- public TreeNode helper(int[] preorder, int preLeft, int preRight,
- int[] inorder, int inLeft, int inRight) {
- // 递归终止条件
- if (inLeft > inRight || preLeft > preRight) return null;
-
- // val 为前序遍历第一个的值，也即是根节点的值
- // idx 为根据根节点的值来找中序遍历的下标
- int idx = inLeft, val = preorder[preLeft];
- TreeNode root = new TreeNode(val);
- for (int i = inLeft; i <= inRight; i++) {
- if (inorder[i] == val) {
- idx = i;
- break;
- }
+ map = new HashMap<>();
+ for (int i = 0; i < inorder.length; i++) { // 用map保存中序序列的数值对应位置
+ map.put(inorder[i], i);
}
- // 根据 idx 来递归找左右子树
- root.left = helper(preorder, preLeft + 1, preLeft + (idx - inLeft),
- inorder, inLeft, idx - 1);
- root.right = helper(preorder, preLeft + (idx - inLeft) + 1, preRight,
- inorder, idx + 1, inRight);
+ return findNode(preorder, 0, preorder.length, inorder, 0, inorder.length); // 前闭后开
+ }
+
+ public TreeNode findNode(int[] preorder, int preBegin, int preEnd, int[] inorder, int inBegin, int inEnd) {
+ // 参数里的范围都是前闭后开
+ if (preBegin >= preEnd || inBegin >= inEnd) { // 不满足左闭右开，说明没有元素，返回空树
+ return null;
+ }
+ int rootIndex = map.get(preorder[preBegin]); // 找到前序遍历的第一个元素在中序遍历中的位置
+ TreeNode root = new TreeNode(inorder[rootIndex]); // 构造结点
+ int lenOfLeft = rootIndex - inBegin; // 保存中序左子树个数，用来确定前序数列的个数
+ root.left = findNode(preorder, preBegin + 1, preBegin + lenOfLeft + 1,
+ inorder, inBegin, rootIndex);
+ root.right = findNode(preorder, preBegin + lenOfLeft + 1, preEnd,
+ inorder, rootIndex + 1, inEnd);
+
return root;
}
}
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/0131.分割回文串.md b/problems/0131.分割回文串.md
index a371864d..64d45853 100644
--- a/problems/0131.分割回文串.md
+++ b/problems/0131.分割回文串.md
@@ -676,5 +676,50 @@ impl Solution {
}
}
```
+
+
+## Scala
+
+```scala
+object Solution {
+
+ import scala.collection.mutable
+
+ def partition(s: String): List[List[String]] = {
+ var result = mutable.ListBuffer[List[String]]()
+ var path = mutable.ListBuffer[String]()
+
+ // 判断字符串是否回文
+ def isPalindrome(start: Int, end: Int): Boolean = {
+ var (left, right) = (start, end)
+ while (left < right) {
+ if (s(left) != s(right)) return false
+ left += 1
+ right -= 1
+ }
+ true
+ }
+
+ // 回溯算法
+ def backtracking(startIndex: Int): Unit = {
+ if (startIndex >= s.size) {
+ result.append(path.toList)
+ return
+ }
+ // 添加循环守卫，如果当前分割是回文子串则进入回溯
+ for (i <- startIndex until s.size if isPalindrome(startIndex, i)) {
+ path.append(s.substring(startIndex, i + 1))
+ backtracking(i + 1)
+ path = path.take(path.size - 1)
+ }
+ }
+
+ backtracking(0)
+ result.toList
+ }
+}
+```
+
+
-----------------------
diff --git a/problems/0141.环形链表.md b/problems/0141.环形链表.md
index ddd83c94..ce90b6c4 100644
--- a/problems/0141.环形链表.md
+++ b/problems/0141.环形链表.md
@@ -7,6 +7,8 @@
# 141. 环形链表
+[力扣题目链接](https://leetcode.cn/problems/linked-list-cycle/submissions/)
+
给定一个链表，判断链表中是否有环。
如果链表中有某个节点，可以通过连续跟踪 next 指针再次到达，则链表中存在环。 为了表示给定链表中的环，我们使用整数 pos 来表示链表尾连接到链表中的位置（索引从 0 开始）。 如果 pos 是 -1，则在该链表中没有环。注意：pos 不作为参数进行传递，仅仅是为了标识链表的实际情况。
@@ -103,7 +105,7 @@ class Solution:
return False
```
-## Go
+### Go
```go
func hasCycle(head *ListNode) bool {
@@ -139,6 +141,23 @@ var hasCycle = function(head) {
};
```
+### TypeScript
+
+```typescript
+function hasCycle(head: ListNode | null): boolean {
+ let slowNode: ListNode | null = head,
+ fastNode: ListNode | null = head;
+ while (fastNode !== null && fastNode.next !== null) {
+ slowNode = slowNode!.next;
+ fastNode = fastNode.next.next;
+ if (slowNode === fastNode) return true;
+ }
+ return false;
+};
+```
+
+
+
-----------------------
diff --git a/problems/0142.环形链表II.md b/problems/0142.环形链表II.md
index 6b7c7e66..658bd868 100644
--- a/problems/0142.环形链表II.md
+++ b/problems/0142.环形链表II.md
@@ -301,13 +301,13 @@ function detectCycle(head: ListNode | null): ListNode | null {
let slowNode: ListNode | null = head,
fastNode: ListNode | null = head;
while (fastNode !== null && fastNode.next !== null) {
- slowNode = (slowNode as ListNode).next;
+ slowNode = slowNode!.next;
fastNode = fastNode.next.next;
if (slowNode === fastNode) {
slowNode = head;
while (slowNode !== fastNode) {
- slowNode = (slowNode as ListNode).next;
- fastNode = (fastNode as ListNode).next;
+ slowNode = slowNode!.next;
+ fastNode = fastNode!.next;
}
return slowNode;
}
diff --git a/problems/0143.重排链表.md b/problems/0143.重排链表.md
index 790bcb48..c60fc0f9 100644
--- a/problems/0143.重排链表.md
+++ b/problems/0143.重排链表.md
@@ -6,6 +6,8 @@
# 143.重排链表
+[力扣题目链接](https://leetcode.cn/problems/reorder-list/submissions/)
+
## 思路
@@ -465,7 +467,81 @@ var reorderList = function(head, s = [], tmp) {
}
```
+### TypeScript
+
+> 辅助数组法：
+
+```typescript
+function reorderList(head: ListNode | null): void {
+ if (head === null) return;
+ const helperArr: ListNode[] = [];
+ let curNode: ListNode | null = head;
+ while (curNode !== null) {
+ helperArr.push(curNode);
+ curNode = curNode.next;
+ }
+ let node: ListNode = head;
+ let left: number = 1,
+ right: number = helperArr.length - 1;
+ let count: number = 0;
+ while (left <= right) {
+ if (count % 2 === 0) {
+ node.next = helperArr[right--];
+ } else {
+ node.next = helperArr[left++];
+ }
+ count++;
+ node = node.next;
+ }
+ node.next = null;
+};
+```
+
+> 分割链表法：
+
+```typescript
+function reorderList(head: ListNode | null): void {
+ if (head === null || head.next === null) return;
+ let fastNode: ListNode = head,
+ slowNode: ListNode = head;
+ while (fastNode.next !== null && fastNode.next.next !== null) {
+ slowNode = slowNode.next!;
+ fastNode = fastNode.next.next;
+ }
+ let head1: ListNode | null = head;
+ // 反转后半部分链表
+ let head2: ListNode | null = reverseList(slowNode.next);
+ // 分割链表
+ slowNode.next = null;
+ /**
+ 直接在head1链表上进行插入
+ head1 链表长度一定大于或等于head2,
+ 因此在下面的循环中，只要head2不为null, head1 一定不为null
+ */
+ while (head2 !== null) {
+ const tempNode1: ListNode | null = head1!.next,
+ tempNode2: ListNode | null = head2.next;
+ head1!.next = head2;
+ head2.next = tempNode1;
+ head1 = tempNode1;
+ head2 = tempNode2;
+ }
+};
+function reverseList(head: ListNode | null): ListNode | null {
+ let curNode: ListNode | null = head,
+ preNode: ListNode | null = null;
+ while (curNode !== null) {
+ const tempNode: ListNode | null = curNode.next;
+ curNode.next = preNode;
+ preNode = curNode;
+ curNode = tempNode;
+ }
+ return preNode;
+}
+```
+
### C
+
方法三：反转链表
```c
//翻转链表
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/0209.长度最小的子数组.md b/problems/0209.长度最小的子数组.md
index e2eb378e..090e73f0 100644
--- a/problems/0209.长度最小的子数组.md
+++ b/problems/0209.长度最小的子数组.md
@@ -465,6 +465,27 @@ object Solution {
}
}
```
-
+C#:
+```csharp
+public class Solution {
+ public int MinSubArrayLen(int s, int[] nums) {
+ int n = nums.Length;
+ int ans = int.MaxValue;
+ int start = 0, end = 0;
+ int sum = 0;
+ while (end < n) {
+ sum += nums[end];
+ while (sum >= s)
+ {
+ ans = Math.Min(ans, end - start + 1);
+ sum -= nums[start];
+ start++;
+ }
+ end++;
+ }
+ return ans == int.MaxValue ? 0 : ans;
+ }
+}
+```
-----------------------
diff --git a/problems/0216.组合总和III.md b/problems/0216.组合总和III.md
index 0ace0fd5..2c1bf717 100644
--- a/problems/0216.组合总和III.md
+++ b/problems/0216.组合总和III.md
@@ -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
+ }
+}
+```
+
+
-----------------------
diff --git a/problems/0234.回文链表.md b/problems/0234.回文链表.md
index f591fcef..eeee6fa5 100644
--- a/problems/0234.回文链表.md
+++ b/problems/0234.回文链表.md
@@ -273,7 +273,7 @@ class Solution:
return pre
```
-## Go
+### Go
```go
@@ -319,6 +319,63 @@ var isPalindrome = function(head) {
};
```
+### TypeScript
+
+> 数组模拟
+
+```typescript
+function isPalindrome(head: ListNode | null): boolean {
+ const helperArr: number[] = [];
+ let curNode: ListNode | null = head;
+ while (curNode !== null) {
+ helperArr.push(curNode.val);
+ curNode = curNode.next;
+ }
+ let left: number = 0,
+ right: number = helperArr.length - 1;
+ while (left < right) {
+ if (helperArr[left++] !== helperArr[right--]) return false;
+ }
+ return true;
+};
+```
+
+> 反转后半部分链表
+
+```typescript
+function isPalindrome(head: ListNode | null): boolean {
+ if (head === null || head.next === null) return true;
+ let fastNode: ListNode | null = head,
+ slowNode: ListNode = head,
+ preNode: ListNode = head;
+ while (fastNode !== null && fastNode.next !== null) {
+ preNode = slowNode;
+ slowNode = slowNode.next!;
+ fastNode = fastNode.next.next;
+ }
+ preNode.next = null;
+ let cur1: ListNode | null = head;
+ let cur2: ListNode | null = reverseList(slowNode);
+ while (cur1 !== null) {
+ if (cur1.val !== cur2!.val) return false;
+ cur1 = cur1.next;
+ cur2 = cur2!.next;
+ }
+ return true;
+};
+function reverseList(head: ListNode | null): ListNode | null {
+ let curNode: ListNode | null = head,
+ preNode: ListNode | null = null;
+ while (curNode !== null) {
+ let tempNode: ListNode | null = curNode.next;
+ curNode.next = preNode;
+ preNode = curNode;
+ curNode = tempNode;
+ }
+ return preNode;
+}
+```
+
-----------------------
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/0344.反转字符串.md b/problems/0344.反转字符串.md
index aba6e2f3..90dbcc0e 100644
--- a/problems/0344.反转字符串.md
+++ b/problems/0344.反转字符串.md
@@ -190,13 +190,13 @@ javaScript:
* @return {void} Do not return anything, modify s in-place instead.
*/
var reverseString = function(s) {
- return s.reverse();
+ //Do not return anything, modify s in-place instead.
+ reverse(s)
};
-var reverseString = function(s) {
+var reverse = function(s) {
let l = -1, r = s.length;
while(++l < --r) [s[l], s[r]] = [s[r], s[l]];
- return s;
};
```
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/0707.设计链表.md b/problems/0707.设计链表.md
index 6ee11eef..42e3fc09 100644
--- a/problems/0707.设计链表.md
+++ b/problems/0707.设计链表.md
@@ -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);
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 19319205..59816eb3 100644
--- a/problems/0977.有序数组的平方.md
+++ b/problems/0977.有序数组的平方.md
@@ -421,6 +421,24 @@ 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;
+ }
+}
+```
-----------------------
diff --git a/problems/背包理论基础01背包-1.md b/problems/背包理论基础01背包-1.md
index a40a92ab..e24824b9 100644
--- a/problems/背包理论基础01背包-1.md
+++ b/problems/背包理论基础01背包-1.md
@@ -356,9 +356,13 @@ 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++ {
- dp[i][j] = max(dp[i-1][j], dp[i-1][j-weight[i]]+value[i])
- }
+ 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]
}