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Commit af239ce

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Merge branch 'master' of github.com:youngyangyang04/leetcode-master
2 parents f5f5f5a + e4d638f commit af239ce

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‎problems/0001.两数之和.md

Lines changed: 43 additions & 0 deletions
Original file line numberDiff line numberDiff line change
@@ -275,5 +275,48 @@ class Solution {
275275
}
276276
```
277277

278+
Scala:
279+
```scala
280+
object Solution {
281+
// 导入包
282+
import scala.collection.mutable
283+
def twoSum(nums: Array[Int], target: Int): Array[Int] = {
284+
// key存储值,value存储下标
285+
val map = new mutable.HashMap[Int, Int]()
286+
for (i <- nums.indices) {
287+
val tmp = target - nums(i) // 计算差值
288+
// 如果这个差值存在于map,则说明找到了结果
289+
if (map.contains(tmp)) {
290+
return Array(map.get(tmp).get, i)
291+
}
292+
// 如果不包含把当前值与其下标放到map
293+
map.put(nums(i), i)
294+
}
295+
// 如果没有找到直接返回一个空的数组,return关键字可以省略
296+
new Array[Int](2)
297+
}
298+
}
299+
```
300+
301+
C#:
302+
```csharp
303+
public class Solution {
304+
public int[] TwoSum(int[] nums, int target) {
305+
Dictionary<int ,int> dic= new Dictionary<int,int>();
306+
for(int i=0;i<nums.Length;i++){
307+
int imp= target-nums[i];
308+
if(dic.ContainsKey(imp)&&dic[imp]!=i){
309+
return new int[]{i, dic[imp]};
310+
}
311+
if(!dic.ContainsKey(nums[i])){
312+
dic.Add(nums[i],i);
313+
}
314+
}
315+
return new int[]{0, 0};
316+
}
317+
}
318+
```
319+
320+
278321
-----------------------
279322
<div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

‎problems/0015.三数之和.md

Lines changed: 44 additions & 1 deletion
Original file line numberDiff line numberDiff line change
@@ -616,6 +616,49 @@ public class Solution
616616
}
617617
}
618618
```
619-
619+
Scala:
620+
```scala
621+
object Solution {
622+
// 导包
623+
import scala.collection.mutable.ListBuffer
624+
import scala.util.control.Breaks.{break, breakable}
625+
626+
def threeSum(nums: Array[Int]): List[List[Int]] = {
627+
// 定义结果集,最后需要转换为List
628+
val res = ListBuffer[List[Int]]()
629+
val nums_tmp = nums.sorted // 对nums进行排序
630+
for (i <- nums_tmp.indices) {
631+
// 如果要排的第一个数字大于0,直接返回结果
632+
if (nums_tmp(i) > 0) {
633+
return res.toList
634+
}
635+
// 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
636+
breakable {
637+
if (i > 0 && nums_tmp(i) == nums_tmp(i - 1)) {
638+
break
639+
} else {
640+
var left = i + 1
641+
var right = nums_tmp.length - 1
642+
while (left < right) {
643+
var sum = nums_tmp(i) + nums_tmp(left) + nums_tmp(right) // 求三数之和
644+
if (sum < 0) left += 1
645+
else if (sum > 0) right -= 1
646+
else {
647+
res += List(nums_tmp(i), nums_tmp(left), nums_tmp(right)) // 如果等于0 添加进结果集
648+
// 为了避免重复,对left和right进行移动
649+
while (left < right && nums_tmp(left) == nums_tmp(left + 1)) left += 1
650+
while (left < right && nums_tmp(right) == nums_tmp(right - 1)) right -= 1
651+
left += 1
652+
right -= 1
653+
}
654+
}
655+
}
656+
}
657+
}
658+
// 最终返回需要转换为List,return关键字可以省略
659+
res.toList
660+
}
661+
}
662+
```
620663
-----------------------
621664
<div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

‎problems/0018.四数之和.md

Lines changed: 44 additions & 1 deletion
Original file line numberDiff line numberDiff line change
@@ -522,6 +522,49 @@ public class Solution
522522
}
523523
}
524524
```
525-
525+
Scala:
526+
```scala
527+
object Solution {
528+
// 导包
529+
import scala.collection.mutable.ListBuffer
530+
import scala.util.control.Breaks.{break, breakable}
531+
def fourSum(nums: Array[Int], target: Int): List[List[Int]] = {
532+
val res = ListBuffer[List[Int]]()
533+
val nums_tmp = nums.sorted // 先排序
534+
for (i <- nums_tmp.indices) {
535+
breakable {
536+
if (i > 0 && nums_tmp(i) == nums_tmp(i - 1)) {
537+
break // 如果该值和上次的值相同,跳过本次循环,相当于continue
538+
} else {
539+
for (j <- i + 1 until nums_tmp.length) {
540+
breakable {
541+
if (j > i + 1 && nums_tmp(j) == nums_tmp(j - 1)) {
542+
break // 同上
543+
} else {
544+
// 双指针
545+
var (left, right) = (j + 1, nums_tmp.length - 1)
546+
while (left < right) {
547+
var sum = nums_tmp(i) + nums_tmp(j) + nums_tmp(left) + nums_tmp(right)
548+
if (sum == target) {
549+
// 满足要求,直接加入到集合里面去
550+
res += List(nums_tmp(i), nums_tmp(j), nums_tmp(left), nums_tmp(right))
551+
while (left < right && nums_tmp(left) == nums_tmp(left + 1)) left += 1
552+
while (left < right && nums_tmp(right) == nums_tmp(right - 1)) right -= 1
553+
left += 1
554+
right -= 1
555+
} else if (sum < target) left += 1
556+
else right -= 1
557+
}
558+
}
559+
}
560+
}
561+
}
562+
}
563+
}
564+
// 最终返回的res要转换为List,return关键字可以省略
565+
res.toList
566+
}
567+
}
568+
```
526569
-----------------------
527570
<div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

‎problems/0019.删除链表的倒数第N个节点.md

Lines changed: 24 additions & 2 deletions
Original file line numberDiff line numberDiff line change
@@ -39,7 +39,7 @@
3939

4040
分为如下几步:
4141

42-
* 首先这里我推荐大家使用虚拟头结点,这样方面处理删除实际头结点的逻辑,如果虚拟头结点不清楚,可以看这篇: [链表:听说用虚拟头节点会方便很多?](https://programmercarl.com/0203.移除链表元素.html)
42+
* 首先这里我推荐大家使用虚拟头结点,这样方便处理删除实际头结点的逻辑,如果虚拟头结点不清楚,可以看这篇: [链表:听说用虚拟头节点会方便很多?](https://programmercarl.com/0203.移除链表元素.html)
4343

4444
* 定义fast指针和slow指针,初始值为虚拟头结点,如图:
4545

@@ -289,6 +289,28 @@ func removeNthFromEnd(_ head: ListNode?, _ n: Int) -> ListNode? {
289289
return dummyHead.next
290290
}
291291
```
292-
292+
Scala:
293+
```scala
294+
object Solution {
295+
def removeNthFromEnd(head: ListNode, n: Int): ListNode = {
296+
val dummy = new ListNode(-1, head) // 定义虚拟头节点
297+
var fast = head // 快指针从头开始走
298+
var slow = dummy // 慢指针从虚拟头开始头
299+
// 因为参数 n 是不可变量,所以不能使用 while(n>0){n-=1}的方式
300+
for (i <- 0 until n) {
301+
fast = fast.next
302+
}
303+
// 快指针和满指针一起走,直到fast走到null
304+
while (fast != null) {
305+
slow = slow.next
306+
fast = fast.next
307+
}
308+
// 删除slow的下一个节点
309+
slow.next = slow.next.next
310+
// 返回虚拟头节点的下一个
311+
dummy.next
312+
}
313+
}
314+
```
293315
-----------------------
294316
<div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

‎problems/0020.有效的括号.md

Lines changed: 22 additions & 1 deletion
Original file line numberDiff line numberDiff line change
@@ -400,6 +400,27 @@ bool isValid(char * s){
400400
return !stackTop;
401401
}
402402
```
403-
403+
Scala:
404+
```scala
405+
object Solution {
406+
import scala.collection.mutable
407+
def isValid(s: String): Boolean = {
408+
if(s.length % 2 != 0) return false // 如果字符串长度是奇数直接返回false
409+
val stack = mutable.Stack[Char]()
410+
// 循环遍历字符串
411+
for (i <- s.indices) {
412+
val c = s(i)
413+
if (c == '(' || c == '[' || c == '{') stack.push(c)
414+
else if(stack.isEmpty) return false // 如果没有(、[、{则直接返回false
415+
// 以下三种情况,不满足则直接返回false
416+
else if(c==')' && stack.pop() != '(') return false
417+
else if(c==']' && stack.pop() != '[') return false
418+
else if(c=='}' && stack.pop() != '{') return false
419+
}
420+
// 如果为空则正确匹配,否则还有余孽就不匹配
421+
stack.isEmpty
422+
}
423+
}
424+
```
404425
-----------------------
405426
<div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

‎problems/0024.两两交换链表中的节点.md

Lines changed: 23 additions & 1 deletion
Original file line numberDiff line numberDiff line change
@@ -311,7 +311,29 @@ func swapPairs(_ head: ListNode?) -> ListNode? {
311311
return dummyHead.next
312312
}
313313
```
314-
314+
Scala:
315+
```scala
316+
// 虚拟头节点
317+
object Solution {
318+
def swapPairs(head: ListNode): ListNode = {
319+
var dummy = new ListNode(0, head) // 虚拟头节点
320+
var pre = dummy
321+
var cur = head
322+
// 当pre的下一个和下下个都不为空,才进行两两转换
323+
while (pre.next != null && pre.next.next != null) {
324+
var tmp: ListNode = cur.next.next // 缓存下一次要进行转换的第一个节点
325+
pre.next = cur.next // 步骤一
326+
cur.next.next = cur // 步骤二
327+
cur.next = tmp // 步骤三
328+
// 下面是准备下一轮的交换
329+
pre = cur
330+
cur = tmp
331+
}
332+
// 最终返回dummy虚拟头节点的下一个,return可以省略
333+
dummy.next
334+
}
335+
}
336+
```
315337

316338
-----------------------
317339
<div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

‎problems/0027.移除元素.md

Lines changed: 16 additions & 2 deletions
Original file line numberDiff line numberDiff line change
@@ -93,7 +93,7 @@ public:
9393

9494
**双指针法(快慢指针法)在数组和链表的操作中是非常常见的,很多考察数组、链表、字符串等操作的面试题,都使用双指针法。**
9595

96-
后序都会一一介绍到,本题代码如下:
96+
后续都会一一介绍到,本题代码如下:
9797

9898
```CPP
9999
// 时间复杂度:O(n)
@@ -338,6 +338,20 @@ int removeElement(int* nums, int numsSize, int val){
338338
return slow;
339339
}
340340
```
341-
341+
Scala:
342+
```scala
343+
object Solution {
344+
def removeElement(nums: Array[Int], `val`: Int): Int = {
345+
var slow = 0
346+
for (fast <- 0 until nums.length) {
347+
if (`val` != nums(fast)) {
348+
nums(slow) = nums(fast)
349+
slow += 1
350+
}
351+
}
352+
slow
353+
}
354+
}
355+
```
342356
-----------------------
343357
<div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

‎problems/0028.实现strStr.md

Lines changed: 75 additions & 0 deletions
Original file line numberDiff line numberDiff line change
@@ -1166,5 +1166,80 @@ func strStr(_ haystack: String, _ needle: String) -> Int {
11661166

11671167
```
11681168

1169+
PHP:
1170+
1171+
> 前缀表统一减一
1172+
```php
1173+
function strStr($haystack, $needle) {
1174+
if (strlen($needle) == 0) return 0;
1175+
$next= [];
1176+
$this->getNext($next,$needle);
1177+
1178+
$j = -1;
1179+
for ($i = 0;$i < strlen($haystack); $i++) { // 注意i就从0开始
1180+
while($j >= 0 && $haystack[$i] != $needle[$j + 1]) {
1181+
$j = $next[$j];
1182+
}
1183+
if ($haystack[$i] == $needle[$j + 1]) {
1184+
$j++;
1185+
}
1186+
if ($j == (strlen($needle) - 1) ) {
1187+
return ($i - strlen($needle) + 1);
1188+
}
1189+
}
1190+
return -1;
1191+
}
1192+
1193+
function getNext(&$next, $s){
1194+
$j = -1;
1195+
$next[0] = $j;
1196+
for($i = 1; $i < strlen($s); $i++) { // 注意i从1开始
1197+
while ($j >= 0 && $s[$i] != $s[$j + 1]) {
1198+
$j = $next[$j];
1199+
}
1200+
if ($s[$i] == $s[$j + 1]) {
1201+
$j++;
1202+
}
1203+
$next[$i] = $j;
1204+
}
1205+
}
1206+
```
1207+
1208+
> 前缀表统一不减一
1209+
```php
1210+
function strStr($haystack, $needle) {
1211+
if (strlen($needle) == 0) return 0;
1212+
$next= [];
1213+
$this->getNext($next,$needle);
1214+
1215+
$j = 0;
1216+
for ($i = 0;$i < strlen($haystack); $i++) { // 注意i就从0开始
1217+
while($j > 0 && $haystack[$i] != $needle[$j]) {
1218+
$j = $next[$j-1];
1219+
}
1220+
if ($haystack[$i] == $needle[$j]) {
1221+
$j++;
1222+
}
1223+
if ($j == strlen($needle)) {
1224+
return ($i - strlen($needle) + 1);
1225+
}
1226+
}
1227+
return -1;
1228+
}
1229+
1230+
function getNext(&$next, $s){
1231+
$j = 0;
1232+
$next[0] = $j;
1233+
for($i = 1; $i < strlen($s); $i++) { // 注意i从1开始
1234+
while ($j > 0 && $s[$i] != $s[$j]) {
1235+
$j = $next[$j-1];
1236+
}
1237+
if ($s[$i] == $s[$j]) {
1238+
$j++;
1239+
}
1240+
$next[$i] = $j;
1241+
}
1242+
}
1243+
```
11691244
-----------------------
11701245
<div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

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