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Merge branch 'youngyangyang04:master' into master

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chujia dang
2022-06-07 17:32:22 +08:00
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parent 9ccacf5ed5 87abfa1664
commit d3e97062bd
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25
problems/0001.两数之和.md
View File

@ -274,6 +274,30 @@ class Solution {
}
}
```
Scala:
```scala
object Solution {
// 导入包
import scala.collection.mutable
def twoSum(nums: Array[Int], target: Int): Array[Int] = {
// key存储值，value存储下标
val map = new mutable.HashMap[Int, Int]()
for (i <- nums.indices) {
val tmp = target - nums(i) // 计算差值
// 如果这个差值存在于map，则说明找到了结果
if (map.contains(tmp)) {
return Array(map.get(tmp).get, i)
}
// 如果不包含把当前值与其下标放到map
map.put(nums(i), i)
}
// 如果没有找到直接返回一个空的数组，return关键字可以省略
new Array[Int](2)
}
}
```
C#:
```csharp
public class Solution {
@ -293,5 +317,6 @@ public class Solution {
}
```
-----------------------
<div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

43
problems/0015.三数之和.md
View File

@ -616,6 +616,49 @@ public class Solution
}
}
```
Scala:
```scala
object Solution {
// 导包
import scala.collection.mutable.ListBuffer
import scala.util.control.Breaks.{break, breakable}
def threeSum(nums: Array[Int]): List[List[Int]] = {
// 定义结果集，最后需要转换为List
val res = ListBuffer[List[Int]]()
val nums_tmp = nums.sorted // 对nums进行排序
for (i <- nums_tmp.indices) {
// 如果要排的第一个数字大于0，直接返回结果
if (nums_tmp(i) > 0) {
return res.toList
}
// 如果i大于0并且和前一个数字重复，则跳过本次循环，相当于continue
breakable {
if (i > 0 && nums_tmp(i) == nums_tmp(i - 1)) {
break
} else {
var left = i + 1
var right = nums_tmp.length - 1
while (left < right) {
var sum = nums_tmp(i) + nums_tmp(left) + nums_tmp(right) // 求三数之和
if (sum < 0) left += 1
else if (sum > 0) right -= 1
else {
res += List(nums_tmp(i), nums_tmp(left), nums_tmp(right)) // 如果等于0 添加进结果集
// 为了避免重复，对left和right进行移动
while (left < right && nums_tmp(left) == nums_tmp(left + 1)) left += 1
while (left < right && nums_tmp(right) == nums_tmp(right - 1)) right -= 1
left += 1
right -= 1
}
}
}
}
}
// 最终返回需要转换为List，return关键字可以省略
res.toList
}
}
```
-----------------------
<div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

45
problems/0018.四数之和.md
View File

@ -522,6 +522,49 @@ public class Solution
}
}
```
Scala:
```scala
object Solution {
// 导包
import scala.collection.mutable.ListBuffer
import scala.util.control.Breaks.{break, breakable}
def fourSum(nums: Array[Int], target: Int): List[List[Int]] = {
val res = ListBuffer[List[Int]]()
val nums_tmp = nums.sorted // 先排序
for (i <- nums_tmp.indices) {
breakable {
if (i > 0 && nums_tmp(i) == nums_tmp(i - 1)) {
break // 如果该值和上次的值相同，跳过本次循环，相当于continue
} else {
for (j <- i + 1 until nums_tmp.length) {
breakable {
if (j > i + 1 && nums_tmp(j) == nums_tmp(j - 1)) {
break // 同上
} else {
// 双指针
var (left, right) = (j + 1, nums_tmp.length - 1)
while (left < right) {
var sum = nums_tmp(i) + nums_tmp(j) + nums_tmp(left) + nums_tmp(right)
if (sum == target) {
// 满足要求，直接加入到集合里面去
res += List(nums_tmp(i), nums_tmp(j), nums_tmp(left), nums_tmp(right))
while (left < right && nums_tmp(left) == nums_tmp(left + 1)) left += 1
while (left < right && nums_tmp(right) == nums_tmp(right - 1)) right -= 1
left += 1
right -= 1
} else if (sum < target) left += 1
else right -= 1
}
}
}
}
}
}
}
// 最终返回的res要转换为List，return关键字可以省略
res.toList
}
}
```
-----------------------
<div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

23
problems/0020.有效的括号.md
View File

@ -400,6 +400,27 @@ bool isValid(char * s){
return !stackTop;
}
```
Scala:
```scala
object Solution {
import scala.collection.mutable
def isValid(s: String): Boolean = {
if(s.length % 2 != 0) return false // 如果字符串长度是奇数直接返回false
val stack = mutable.Stack[Char]()
// 循环遍历字符串
for (i <- s.indices) {
val c = s(i)
if (c == '(' || c == '[' || c == '{') stack.push(c)
else if(stack.isEmpty) return false // 如果没有(、[、{则直接返回false
// 以下三种情况，不满足则直接返回false
else if(c==')' && stack.pop() != '(') return false
else if(c==']' && stack.pop() != '[') return false
else if(c=='}' && stack.pop() != '{') return false
}
// 如果为空则正确匹配，否则还有余孽就不匹配
stack.isEmpty
}
}
```
-----------------------
<div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

75
problems/0028.实现strStr.md
View File

@ -1166,5 +1166,80 @@ func strStr(_ haystack: String, _ needle: String) -> Int {
```
PHP：
> 前缀表统一减一
```php
function strStr($haystack, $needle) {
if (strlen($needle) == 0) return 0;
$next= [];
$this->getNext($next,$needle);
$j = -1;
for ($i = 0;$i < strlen($haystack); $i++) { // 注意i就从0开始
while($j >= 0 && $haystack[$i] != $needle[$j + 1]) {
$j = $next[$j];
}
if ($haystack[$i] == $needle[$j + 1]) {
$j++;
}
if ($j == (strlen($needle) - 1) ) {
return ($i - strlen($needle) + 1);
}
}
return -1;
}
function getNext(&$next, $s){
$j = -1;
$next[0] = $j;
for($i = 1; $i < strlen($s); $i++) { // 注意i从1开始
while ($j >= 0 && $s[$i] != $s[$j + 1]) {
$j = $next[$j];
}
if ($s[$i] == $s[$j + 1]) {
$j++;
}
$next[$i] = $j;
}
}
```
> 前缀表统一不减一
```php
function strStr($haystack, $needle) {
if (strlen($needle) == 0) return 0;
$next= [];
$this->getNext($next,$needle);
$j = 0;
for ($i = 0;$i < strlen($haystack); $i++) { // 注意i就从0开始
while($j > 0 && $haystack[$i] != $needle[$j]) {
$j = $next[$j-1];
}
if ($haystack[$i] == $needle[$j]) {
$j++;
}
if ($j == strlen($needle)) {
return ($i - strlen($needle) + 1);
}
}
return -1;
}
function getNext(&$next, $s){
$j = 0;
$next[0] = $j;
for($i = 1; $i < strlen($s); $i++) { // 注意i从1开始
while ($j > 0 && $s[$i] != $s[$j]) {
$j = $next[$j-1];
}
if ($s[$i] == $s[$j]) {
$j++;
}
$next[$i] = $j;
}
}
```
-----------------------
<div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

2
problems/0056.合并区间.md
View File

@ -96,7 +96,7 @@ public:
vector<vector<int>> merge(vector<vector<int>>& intervals) {
vector<vector<int>> result;
if (intervals.size() == 0) return result;
// 排序的参数使用了lamda表达式
// 排序的参数使用了lambda表达式
sort(intervals.begin(), intervals.end(), [](const vector<int>& a, const vector<int>& b){return a[0] < b[0];});
result.push_back(intervals[0]);

6
problems/0063.不同路径II.md
View File

@ -66,7 +66,7 @@ dp[i][j] ：表示从（0 ，0）出发，到(i, j) 有dp[i][j]条不同的路
所以代码为：
```
```cpp
if (obstacleGrid[i][j] == 0) { // 当(i, j)没有障碍的时候，再推导dp[i][j]
dp[i][j] = dp[i - 1][j] + dp[i][j - 1];
}
@ -76,7 +76,7 @@ if (obstacleGrid[i][j] == 0) { // 当(i, j)没有障碍的时候，再推导dp[i
在[62.不同路径](https://programmercarl.com/0062.不同路径.html)不同路径中我们给出如下的初始化：
```
```cpp
vector<vector<int>> dp(m, vector<int>(n, 0)); // 初始值为0
for (int i = 0; i < m; i++) dp[i][0] = 1;
for (int j = 0; j < n; j++) dp[0][j] = 1;
@ -138,6 +138,8 @@ public:
int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) {
int m = obstacleGrid.size();
int n = obstacleGrid[0].size();
if (obstacleGrid[m - 1][n - 1] == 1 || obstacleGrid[0][0] == 1) //如果在起点或终点出现了障碍，直接返回0
return 0;
vector<vector<int>> dp(m, vector<int>(n, 0));
for (int i = 0; i < m && obstacleGrid[i][0] == 0; i++) dp[i][0] = 1;
for (int j = 0; j < n && obstacleGrid[0][j] == 0; j++) dp[0][j] = 1;

59
problems/0102.二叉树的层序遍历.md
View File

@ -319,6 +319,36 @@ func levelOrder(_ root: TreeNode?) -> [[Int]] {
}
```
Rust:
```rust
pub fn level_order(root: Option<Rc<RefCell<TreeNode>>>) -> Vec<Vec<i32>> {
let mut ans = Vec::new();
let mut stack = Vec::new();
if root.is_none(){
return ans;
}
stack.push(root.unwrap());
while stack.is_empty()!= true{
let num = stack.len();
let mut level = Vec::new();
for _i in 0..num{
let tmp = stack.remove(0);
level.push(tmp.borrow_mut().val);
if tmp.borrow_mut().left.is_some(){
stack.push(tmp.borrow_mut().left.take().unwrap());
}
if tmp.borrow_mut().right.is_some(){
stack.push(tmp.borrow_mut().right.take().unwrap());
}
}
ans.push(level);
}
ans
}
```
**此时我们就掌握了二叉树的层序遍历了，那么如下九道力扣上的题目，只需要修改模板的两三行代码（不能再多了），便可打倒！**
@ -548,6 +578,35 @@ func levelOrderBottom(_ root: TreeNode?) -> [[Int]] {
}
```
Rust:
```rust
pub fn level_order(root: Option<Rc<RefCell<TreeNode>>>) -> Vec<Vec<i32>> {
let mut ans = Vec::new();
let mut stack = Vec::new();
if root.is_none(){
return ans;
}
stack.push(root.unwrap());
while stack.is_empty()!= true{
let num = stack.len();
let mut level = Vec::new();
for _i in 0..num{
let tmp = stack.remove(0);
level.push(tmp.borrow_mut().val);
if tmp.borrow_mut().left.is_some(){
stack.push(tmp.borrow_mut().left.take().unwrap());
}
if tmp.borrow_mut().right.is_some(){
stack.push(tmp.borrow_mut().right.take().unwrap());
}
}
ans.push(level);
}
ans
}
```
# 199.二叉树的右视图
[力扣题目链接](https://leetcode-cn.com/problems/binary-tree-right-side-view/)

27
problems/0104.二叉树的最大深度.md
View File

@ -192,6 +192,33 @@ public:
};
```
rust:
```rust
impl Solution {
pub fn max_depth(root: Option<Rc<RefCell<TreeNode>>>) -> i32 {
if root.is_none(){
return 0;
}
let mut max_depth: i32 = 0;
let mut stack = vec![root.unwrap()];
while !stack.is_empty() {
let num = stack.len();
for _i in 0..num{
let top = stack.remove(0);
if top.borrow_mut().left.is_some(){
stack.push(top.borrow_mut().left.take().unwrap());
}
if top.borrow_mut().right.is_some(){
stack.push(top.borrow_mut().right.take().unwrap());
}
}
max_depth+=1;
}
max_depth
}
```
那么我们可以顺便解决一下n叉树的最大深度问题
# 559.n叉树的最大深度

64
problems/0111.二叉树的最小深度.md
View File

@ -488,5 +488,69 @@ func minDepth(_ root: TreeNode?) -> Int {
}
```
rust:
```rust
impl Solution {
pub fn min_depth(root: Option<Rc<RefCell<TreeNode>>>) -> i32 {
return Solution::bfs(root)
}
// 递归
pub fn dfs(node: Option<Rc<RefCell<TreeNode>>>) -> i32{
if node.is_none(){
return 0;
}
let parent = node.unwrap();
let left_child = parent.borrow_mut().left.take();
let right_child = parent.borrow_mut().right.take();
if left_child.is_none() && right_child.is_none(){
return 1;
}
let mut min_depth = i32::MAX;
if left_child.is_some(){
let left_depth = Solution::dfs(left_child);
if left_depth <= min_depth{
min_depth = left_depth
}
}
if right_child.is_some(){
let right_depth = Solution::dfs(right_child);
if right_depth <= min_depth{
min_depth = right_depth
}
}
min_depth + 1
}
// 迭代
pub fn bfs(node: Option<Rc<RefCell<TreeNode>>>) -> i32{
let mut min_depth = 0;
if node.is_none(){
return min_depth
}
let mut stack = vec![node.unwrap()];
while !stack.is_empty(){
min_depth += 1;
let num = stack.len();
for _i in 0..num{
let top = stack.remove(0);
let left_child = top.borrow_mut().left.take();
let right_child = top.borrow_mut().right.take();
if left_child.is_none() && right_child.is_none(){
return min_depth;
}
if left_child.is_some(){
stack.push(left_child.unwrap());
}
if right_child.is_some(){
stack.push(right_child.unwrap());
}
}
}
min_depth
}
```
-----------------------
<div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

29
problems/0150.逆波兰表达式求值.md
View File

@ -325,6 +325,33 @@ func evalRPN(_ tokens: [String]) -> Int {
return stack.last!
}
```
Scala:
```scala
object Solution {
import scala.collection.mutable
def evalRPN(tokens: Array[String]): Int = {
val stack = mutable.Stack[Int]() // 定义栈
// 抽取运算操作，需要传递x,y，和一个函数
def operator(x: Int, y: Int, f: (Int, Int) => Int): Int = f(x, y)
for (token <- tokens) {
// 模式匹配，匹配不同的操作符做什么样的运算
token match {
// 最后一个参数 _+_，代表x+y，遵循Scala的函数至简原则，以下运算同理
case "+" => stack.push(operator(stack.pop(), stack.pop(), _ + _))
case "-" => stack.push(operator(stack.pop(), stack.pop(), -_ + _))
case "*" => stack.push(operator(stack.pop(), stack.pop(), _ * _))
case "/" => {
var pop1 = stack.pop()
var pop2 = stack.pop()
stack.push(operator(pop2, pop1, _ / _))
}
case _ => stack.push(token.toInt) // 不是运算符就入栈
}
}
// 最后返回栈顶，不需要加return关键字
stack.pop()
}
}
```
-----------------------
<div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

56
problems/0151.翻转字符串里的单词.md
View File

@ -758,7 +758,63 @@ func reverseWord(_ s: inout [Character]) {
}
```
Scala:
```scala
object Solution {
def reverseWords(s: String): String = {
var sb = removeSpace(s) // 移除多余的空格
reverseString(sb, 0, sb.length - 1) // 翻转字符串
reverseEachWord(sb)
sb.mkString
}
// 移除多余的空格
def removeSpace(s: String): Array[Char] = {
var start = 0
var end = s.length - 1
// 移除字符串前面的空格
while (start < s.length && s(start) == ' ') start += 1
// 移除字符串后面的空格
while (end >= 0 && s(end) == ' ') end -= 1
var sb = "" // String
// 当start小于等于end的时候，执行添加操作
while (start <= end) {
var c = s(start)
// 当前字符不等于空，sb的最后一个字符不等于空的时候添加到sb
if (c != ' ' || sb(sb.length - 1) != ' ') {
sb ++= c.toString
}
start += 1 // 指针向右移动
}
sb.toArray
}
// 翻转字符串
def reverseString(s: Array[Char], start: Int, end: Int): Unit = {
var (left, right) = (start, end)
while (left < right) {
var tmp = s(left)
s(left) = s(right)
s(right) = tmp
left += 1
right -= 1
}
}
// 翻转每个单词
def reverseEachWord(s: Array[Char]): Unit = {
var i = 0
while (i < s.length) {
var j = i + 1
// 向后迭代寻找每个单词的坐标
while (j < s.length && s(j) != ' ') j += 1
reverseString(s, i, j - 1) // 翻转每个单词
i = j + 1 // i往后更新
}
}
}
```

21
problems/0203.移除链表元素.md
View File

@ -266,6 +266,27 @@ public ListNode removeElements(ListNode head, int val) {
}
return head;
}
/**
* 不添加虚拟节点and pre Node方式
* 时间复杂度 O(n)
* 空间复杂度 O(1)
* @param head
* @param val
* @return
*/
public ListNode removeElements(ListNode head, int val) {
while(head!=null && head.val==val){
head = head.next;
}
ListNode curr = head;
while(curr!=null){
while(curr.next!=null && curr.next.val == val){
curr.next = curr.next.next;
}
curr = curr.next;
}
return head;
}
```
Python：

83
problems/0225.用队列实现栈.md
View File

@ -815,6 +815,89 @@ class MyStack {
}
}
```
Scala:
使用两个队列模拟栈:
```scala
import scala.collection.mutable
class MyStack() {
val queue1 = new mutable.Queue[Int]()
val queue2 = new mutable.Queue[Int]()
def push(x: Int) {
queue1.enqueue(x)
}
def pop(): Int = {
var size = queue1.size
// 将queue1中的每个元素都移动到queue2
for (i <- 0 until size - 1) {
queue2.enqueue(queue1.dequeue())
}
var res = queue1.dequeue()
// 再将queue2中的每个元素都移动到queue1
while (!queue2.isEmpty) {
queue1.enqueue(queue2.dequeue())
}
res
}
def top(): Int = {
var size = queue1.size
for (i <- 0 until size - 1) {
queue2.enqueue(queue1.dequeue())
}
var res = queue1.dequeue()
while (!queue2.isEmpty) {
queue1.enqueue(queue2.dequeue())
}
// 最终还需要把res送进queue1
queue1.enqueue(res)
res
}
def empty(): Boolean = {
queue1.isEmpty
}
}
```
使用一个队列模拟:
```scala
import scala.collection.mutable
class MyStack() {
val queue = new mutable.Queue[Int]()
def push(x: Int) {
queue.enqueue(x)
}
def pop(): Int = {
var size = queue.size
for (i <- 0 until size - 1) {
queue.enqueue(queue.head) // 把头添到队列最后
queue.dequeue() // 再出队
}
queue.dequeue()
}
def top(): Int = {
var size = queue.size
var res = 0
for (i <- 0 until size) {
queue.enqueue(queue.head) // 把头添到队列最后
res = queue.dequeue() // 再出队
}
res
}
def empty(): Boolean = {
queue.isEmpty
}
}
```
-----------------------
<div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

39
problems/0232.用栈实现队列.md
View File

@ -495,6 +495,45 @@ void myQueueFree(MyQueue* obj) {
obj->stackOutTop = 0;
}
```
Scala:
```scala
class MyQueue() {
import scala.collection.mutable
val stackIn = mutable.Stack[Int]() // 负责出栈
val stackOut = mutable.Stack[Int]() // 负责入栈
// 添加元素
def push(x: Int) {
stackIn.push(x)
}
// 复用代码，如果stackOut为空就把stackIn的所有元素都压入StackOut
def dumpStackIn(): Unit = {
if (!stackOut.isEmpty) return
while (!stackIn.isEmpty) {
stackOut.push(stackIn.pop())
}
}
// 弹出元素
def pop(): Int = {
dumpStackIn()
stackOut.pop()
}
// 获取队头
def peek(): Int = {
dumpStackIn()
val res: Int = stackOut.pop()
stackOut.push(res)
res
}
// 判断是否为空
def empty(): Boolean = {
stackIn.isEmpty && stackOut.isEmpty
}
}
```
-----------------------
<div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

47
problems/0239.滑动窗口最大值.md
View File

@ -630,6 +630,53 @@ func maxSlidingWindow(_ nums: [Int], _ k: Int) -> [Int] {
return result
}
```
Scala:
```scala
import scala.collection.mutable.ArrayBuffer
object Solution {
def maxSlidingWindow(nums: Array[Int], k: Int): Array[Int] = {
var len = nums.length - k + 1 // 滑动窗口长度
var res: Array[Int] = new Array[Int](len) // 声明存储结果的数组
var index = 0 // 结果数组指针
val queue: MyQueue = new MyQueue // 自定义队列
// 将前k个添加到queue
for (i <- 0 until k) {
queue.add(nums(i))
}
res(index) = queue.peek // 第一个滑动窗口的最大值
index += 1
for (i <- k until nums.length) {
queue.poll(nums(i - k)) // 首先移除第i-k个元素
queue.add(nums(i)) // 添加当前数字到队列
res(index) = queue.peek() // 赋值
index+=1
}
// 最终返回res，return关键字可以省略
res
}
}
class MyQueue {
var queue = ArrayBuffer[Int]()
// 移除元素，如果传递进来的跟队头相等，那么移除
def poll(value: Int): Unit = {
if (!queue.isEmpty && queue.head == value) {
queue.remove(0)
}
}
// 添加元素，当队尾大于当前元素就删除
def add(value: Int): Unit = {
while (!queue.isEmpty && value > queue.last) {
queue.remove(queue.length - 1)
}
queue.append(value)
}
def peek(): Int = queue.head
}
```
-----------------------
<div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

16
problems/0344.反转字符串.md
View File

@ -266,6 +266,20 @@ public class Solution
}
}
```
Scala:
```scala
object Solution {
def reverseString(s: Array[Char]): Unit = {
var (left, right) = (0, s.length - 1)
while (left < right) {
var tmp = s(left)
s(left) = s(right)
s(right) = tmp
left += 1
right -= 1
}
}
}
```
-----------------------
<div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

42
problems/0347.前K个高频元素.md
View File

@ -374,7 +374,49 @@ function topKFrequent(nums: number[], k: number): number[] {
};
```
Scala:
解法一: 优先级队列
```scala
object Solution {
import scala.collection.mutable
def topKFrequent(nums: Array[Int], k: Int): Array[Int] = {
val map = mutable.HashMap[Int, Int]()
// 将所有元素都放入Map
for (num <- nums) {
map.put(num, map.getOrElse(num, 0) + 1)
}
// 声明一个优先级队列，在函数柯里化那块需要指明排序方式
var queue = mutable.PriorityQueue[(Int, Int)]()(Ordering.fromLessThan((x, y) => x._2 > y._2))
// 将map里面的元素送入优先级队列
for (elem <- map) {
queue.enqueue(elem)
if(queue.size > k){
queue.dequeue // 如果队列元素大于k个，出队
}
}
// 最终只需要key的Array形式就可以了，return关键字可以省略
queue.map(_._1).toArray
}
}
```
解法二: 相当于一个wordCount程序
```scala
object Solution {
def topKFrequent(nums: Array[Int], k: Int): Array[Int] = {
// 首先将数据变为(x,1)，然后按照x分组，再使用map进行转换(x,sum)，变换为Array
// 再使用sort针对sum进行排序，最后take前k个，再把数据变为x,y,z这种格式
nums.map((_, 1)).groupBy(_._1)
.map {
case (x, arr) => (x, arr.map(_._2).sum)
}
.toArray
.sortWith(_._2 > _._2)
.take(k)
.map(_._1)
}
}
```
-----------------------
<div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

65
problems/0383.赎金信.md
View File

@ -363,6 +363,70 @@ impl Solution {
}
```
Scala:
版本一: 使用数组作为哈希表
```scala
object Solution {
def canConstruct(ransomNote: String, magazine: String): Boolean = {
// 如果magazine的长度小于ransomNote的长度，必然是false
if (magazine.length < ransomNote.length) {
return false
}
// 定义一个数组，存储magazine字符出现的次数
val map: Array[Int] = new Array[Int](26)
// 遍历magazine字符串，对应的字符+=1
for (i <- magazine.indices) {
map(magazine(i) - 'a') += 1
}
// 遍历ransomNote
for (i <- ransomNote.indices) {
if (map(ransomNote(i) - 'a') > 0)
map(ransomNote(i) - 'a') -= 1
else return false
}
// 如果上面没有返回false，直接返回true，关键字return可以省略
true
}
}
```
```scala
object Solution {
import scala.collection.mutable
def canConstruct(ransomNote: String, magazine: String): Boolean = {
// 如果magazine的长度小于ransomNote的长度，必然是false
if (magazine.length < ransomNote.length) {
return false
}
// 定义map，key是字符，value是字符出现的次数
val map = new mutable.HashMap[Char, Int]()
// 遍历magazine，把所有的字符都记录到map里面
for (i <- magazine.indices) {
val tmpChar = magazine(i)
// 如果map包含该字符，那么对应的value++，否则添加该字符
if (map.contains(tmpChar)) {
map.put(tmpChar, map.get(tmpChar).get + 1)
} else {
map.put(tmpChar, 1)
}
}
// 遍历ransomNote
for (i <- ransomNote.indices) {
val tmpChar = ransomNote(i)
// 如果map包含并且该字符的value大于0，则匹配成功，map对应的--，否则直接返回false
if (map.contains(tmpChar) && map.get(tmpChar).get > 0) {
map.put(tmpChar, map.get(tmpChar).get - 1)
} else {
return false
}
}
// 如果上面没有返回false，直接返回true，关键字return可以省略
true
}
}
C#：
```csharp
public bool CanConstruct(string ransomNote, string magazine) {
@ -379,6 +443,7 @@ public bool CanConstruct(string ransomNote, string magazine) {
}
return true;
}
```
-----------------------
<div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

4
problems/0404.左叶子之和.md
View File

@ -336,8 +336,8 @@ var sumOfLeftLeaves = function(root) {
if(node===null){
return 0;
}
let leftValue = sumOfLeftLeaves(node.left);
let rightValue = sumOfLeftLeaves(node.right);
let leftValue = nodesSum(node.left);
let rightValue = nodesSum(node.right);
// 3. 单层递归逻辑
let midValue = 0;
if(node.left&&node.left.left===null&&node.left.right===null){

37
problems/0454.四数相加II.md
View File

@ -318,6 +318,43 @@ impl Solution {
}
```
Scala:
```scala
object Solution {
// 导包
import scala.collection.mutable
def fourSumCount(nums1: Array[Int], nums2: Array[Int], nums3: Array[Int], nums4: Array[Int]): Int = {
// 定义一个HashMap，key存储值，value存储该值出现的次数
val map = new mutable.HashMap[Int, Int]()
// 遍历前两个数组，把他们所有可能的情况都记录到map
for (i <- nums1.indices) {
for (j <- nums2.indices) {
val tmp = nums1(i) + nums2(j)
// 如果包含该值，则对他的key加1，不包含则添加进去
if (map.contains(tmp)) {
map.put(tmp, map.get(tmp).get + 1)
} else {
map.put(tmp, 1)
}
}
}
var res = 0 // 结果变量
// 遍历后两个数组
for (i <- nums3.indices) {
for (j <- nums4.indices) {
val tmp = -(nums3(i) + nums4(j))
// 如果map中存在该值，结果就+=value
if (map.contains(tmp)) {
res += map.get(tmp).get
}
}
}
// 返回最终结果，可以省略关键字return
res
}
}
C#：
```csharp
public int FourSumCount(int[] nums1, int[] nums2, int[] nums3, int[] nums4) {

39
problems/0516.最长回文子序列.md
View File

@ -186,29 +186,28 @@ class Solution:
Go：
```Go
func longestPalindromeSubseq(s string) int {
lenth:=len(s)
dp:=make([][]int,lenth)
for i:=0;i<lenth;i++{
for j:=0;j<lenth;j++{
if dp[i]==nil{
dp[i]=make([]int,lenth)
}
if i==j{
dp[i][j]=1
size := len(s)
max := func(a, b int) int {
if a > b {
return a
}
return b
}
dp := make([][]int, size)
for i := 0; i < size; i++ {
dp[i] = make([]int, size)
dp[i][i] = 1
}
for i := size - 1; i >= 0; i-- {
for j := i + 1; j < size; j++ {
if s[i] == s[j] {
dp[i][j] = dp[i+1][j-1] + 2
} else {
dp[i][j] = max(dp[i][j-1], dp[i+1][j])
}
}
}
for i:=lenth-1;i>=0;i--{
for j:=i+1;j<lenth;j++{
if s[i]==s[j]{
dp[i][j]=dp[i+1][j-1]+2
}else {
dp[i][j]=max(dp[i+1][j],dp[i][j-1])
}
}
}
return dp[0][lenth-1]
return dp[0][size-1]
}
```

41
problems/0541.反转字符串II.md
View File

@ -347,6 +347,47 @@ public class Solution
}
}
```
Scala:
版本一: (正常解法)
```scala
object Solution {
def reverseStr(s: String, k: Int): String = {
val res = s.toCharArray // 转换为Array好处理
for (i <- s.indices by 2 * k) {
// 如果i+k大于了res的长度，则需要全部翻转
if (i + k > res.length) {
reverse(res, i, s.length - 1)
} else {
reverse(res, i, i + k - 1)
}
}
new String(res)
}
// 翻转字符串，从start到end
def reverse(s: Array[Char], start: Int, end: Int): Unit = {
var (left, right) = (start, end)
while (left < right) {
var tmp = s(left)
s(left) = s(right)
s(right) = tmp
left += 1
right -= 1
}
}
}
```
版本二: 首先利用sliding每隔k个进行分割，随后转换为数组，再使用zipWithIndex添加每个数组的索引，紧接着利用map做变换，如果索引%2==0则说明需要翻转，否则原封不动，最后再转换为String
```scala
object Solution {
def reverseStr(s: String, k: Int): String = {
s.sliding(k, k)
.toArray
.zipWithIndex
.map(v => if (v._2 % 2 == 0) v._1.reverse else v._1)
.mkString
}
}
```
-----------------------
<div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

23
problems/0714.买卖股票的最佳时机含手续费（动态规划）.md
View File

@ -200,6 +200,29 @@ const maxProfit = (prices,fee) => {
}
```
TypeScript：
```typescript
function maxProfit(prices: number[], fee: number): number {
/**
dp[i][0]：持有股票
dp[i][1]: 不持有
*/
const length: number = prices.length;
if (length === 0) return 0;
const dp: number[][] = new Array(length).fill(0).map(_ => []);
dp[0][0] = -prices[0];
dp[0][1] = 0;
for (let i = 1; i < length; i++) {
dp[i][0] = Math.max(dp[i - 1][0], dp[i - 1][1] - prices[i]);
dp[i][1] = Math.max(dp[i - 1][1], dp[i - 1][0] + prices[i] - fee);
}
return dp[length - 1][1];
};
```
-----------------------
<div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

7
problems/0739.每日温度.md
View File

@ -34,7 +34,8 @@
那么单调栈的原理是什么呢？为什么时间复杂度是O(n)就可以找到每一个元素的右边第一个比它大的元素位置呢？
单调栈的本质是空间换时间，因为在遍历的过程中需要用一个栈来记录右边第一个比当前元素大的元素，优点是只需要遍历一次。
单调栈的本质是空间换时间，因为在遍历的过程中需要用一个栈来记录右边第一个比当前元素高的元素，优点是只需要遍历一次。
在使用单调栈的时候首先要明确如下几点：
@ -192,7 +193,7 @@ class Solution {
否则的话，可以直接入栈。
注意，单调栈里 加入的元素是 下标。
*/
Stack<Integer>stack=new Stack<>();
Deque<Integer> stack=new LinkedList<>();
stack.push(0);
for(int i=1;i<lens;i++){
@ -216,7 +217,7 @@ class Solution {
public int[] dailyTemperatures(int[] temperatures) {
int lens=temperatures.length;
int []res=new int[lens];
Stack<Integer>stack=new Stack<>();
Deque<Integer> stack=new LinkedList<>();
for(int i=0;i<lens;i++){
while(!stack.isEmpty()&&temperatures[i]>temperatures[stack.peek()]){

23
problems/1047.删除字符串中的所有相邻重复项.md
View File

@ -374,6 +374,27 @@ func removeDuplicates(_ s: String) -> String {
return String(stack)
}
```
Scala:
```scala
object Solution {
import scala.collection.mutable
def removeDuplicates(s: String): String = {
var stack = mutable.Stack[Int]()
var str = "" // 保存最终结果
for (i <- s.indices) {
var tmp = s(i)
// 如果栈非空并且栈顶元素等于当前字符，那么删掉栈顶和字符串最后一个元素
if (!stack.isEmpty && tmp == stack.head) {
str = str.take(str.length - 1)
stack.pop()
} else {
stack.push(tmp)
str += tmp
}
}
str
}
}
```
-----------------------
<div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

56
problems/剑指Offer05.替换空格.md
View File

@ -413,8 +413,62 @@ func replaceSpace(_ s: String) -> String {
}
```
Scala:
方式一: 双指针
```scala
object Solution {
def replaceSpace(s: String): String = {
var count = 0
s.foreach(c => if (c == ' ') count += 1) // 统计空格的数量
val sOldSize = s.length // 旧数组字符串长度
val sNewSize = s.length + count * 2 // 新数组字符串长度
val res = new Array[Char](sNewSize) // 新数组
var index = sNewSize - 1 // 新数组索引
// 逆序遍历
for (i <- (0 until sOldSize).reverse) {
if (s(i) == ' ') {
res(index) = '0'
index -= 1
res(index) = '2'
index -= 1
res(index) = '%'
} else {
res(index) = s(i)
}
index -= 1
}
res.mkString
}
}
```
方式二: 使用一个集合，遇到空格就添加%20
```scala
object Solution {
import scala.collection.mutable.ListBuffer
def replaceSpace(s: String): String = {
val res: ListBuffer[Char] = ListBuffer[Char]()
for (i <- s.indices) {
if (s(i) == ' ') {
res += '%'
res += '2'
res += '0'
}else{
res += s(i)
}
}
res.mkString
}
}
```
方式三: 使用map
```scala
object Solution {
def replaceSpace(s: String): String = {
s.map(c => if(c == ' ') "%20" else c).mkString
}
}
```
-----------------------

50
problems/剑指Offer58-II.左旋转字符串.md
View File

@ -291,6 +291,56 @@ func reverseString(_ s: inout [Character], startIndex: Int, endIndex: Int) {
```
### PHP
```php
function reverseLeftWords($s, $n) {
$this->reverse($s,0,$n-1); //反转区间为前n的子串
$this->reverse($s,$n,strlen($s)-1); //反转区间为n到末尾的子串
$this->reverse($s,0,strlen($s)-1); //反转整个字符串
return $s;
}
// 按指定进行翻转 【array、string都可】
function reverse(&$s, $start, $end) {
for ($i = $start, $j = $end; $i < $j; $i++, $j--) {
$tmp = $s[$i];
$s[$i] = $s[$j];
$s[$j] = $tmp;
}
}
```
Scala:
```scala
object Solution {
def reverseLeftWords(s: String, n: Int): String = {
var str = s.toCharArray // 转换为Array
// abcdefg => ba cdefg
reverseString(str, 0, n - 1)
// ba cdefg => ba gfedc
reverseString(str, n, str.length - 1)
// ba gfedc => cdefgab
reverseString(str, 0, str.length - 1)
// 最终返回，return关键字可以省略
new String(str)
}
// 翻转字符串
def reverseString(s: Array[Char], start: Int, end: Int): Unit = {
var (left, right) = (start, end)
while (left < right) {
var tmp = s(left)
s(left) = s(right)
s(right) = tmp
left += 1
right -= 1
}
}
}
```