Merge branch 'youngyangyang04:master' into master

This commit is contained in:
AronJudge
2022-07-11 11:00:10 +08:00
committed by GitHub
33 changed files with 1026 additions and 79 deletions

View File

@ -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
}
}
```
-----------------------
<div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

View File

@ -254,20 +254,19 @@ TypeScript
```typescript
function swapPairs(head: ListNode | null): ListNode | null {
const dummyHead: ListNode = new ListNode(0, head);
let cur: ListNode = dummyHead;
while(cur.next !== null && cur.next.next !== null) {
const tem: ListNode = cur.next;
const tem1: ListNode = cur.next.next.next;
cur.next = cur.next.next; // step 1
cur.next.next = tem; // step 2
cur.next.next.next = tem1; // step 3
cur = cur.next.next;
}
return dummyHead.next;
}
const dummyNode: ListNode = new ListNode(0, head);
let curNode: ListNode | null = dummyNode;
while (curNode && curNode.next && curNode.next.next) {
let firstNode: ListNode = curNode.next,
secNode: ListNode = curNode.next.next,
thirdNode: ListNode | null = curNode.next.next.next;
curNode.next = secNode;
secNode.next = firstNode;
firstNode.next = thirdNode;
curNode = firstNode;
}
return dummyNode.next;
};
```
Kotlin:

View File

@ -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>

View File

@ -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>

View File

@ -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

View File

@ -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
}
}
```
-----------------------
<div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

View File

@ -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
}
}
```
-----------------------
<div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

View File

@ -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>

View File

@ -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>

View File

@ -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
}
}
```
-----------------------
<div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

View File

@ -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<Integer, Integer> 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<Integer, Integer> 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;
}
}

View File

@ -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>

View File

@ -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
}
}
```
-----------------------
<div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

View File

@ -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;
};
```
-----------------------
<div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

View File

@ -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;
}

View File

@ -6,6 +6,8 @@
# 143.重排链表
[力扣题目链接](https://leetcode.cn/problems/reorder-list/submissions/)
![](https://code-thinking-1253855093.file.myqcloud.com/pics/20210726160122.png)
## 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

View File

@ -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()
}
}
```
-----------------------

View File

@ -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;
}
}
```
-----------------------
<div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

View File

@ -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>

View File

@ -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;
}
```
-----------------------

View File

@ -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
}
}
```
-----------------------

View File

@ -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>

View File

@ -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>

View File

@ -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;
};
```

View File

@ -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>

View File

@ -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
}
}
```
-----------------------

View File

@ -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
}
}
```
-----------------------

View File

@ -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>

View File

@ -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:**

View File

@ -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);

View File

@ -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>

View File

@ -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;
}
}
```
-----------------------
<div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

View File

@ -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]
}