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‎problems/0001.两数之和.md

Lines changed: 44 additions & 1 deletion
Original file line numberDiff line numberDiff line change
@@ -7,7 +7,7 @@
77

88
## 1. 两数之和
99

10-
[力扣题目链接](https://leetcode-cn.com/problems/two-sum/)
10+
[力扣题目链接](https://leetcode.cn/problems/two-sum/)
1111

1212
给定一个整数数组 nums 和一个目标值 target,请你在该数组中找出和为目标值的那 两个 整数,并返回他们的数组下标。
1313

@@ -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/0005.最长回文子串.md

Lines changed: 1 addition & 1 deletion
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@@ -8,7 +8,7 @@
88

99
# 5.最长回文子串
1010

11-
[力扣题目链接](https://leetcode-cn.com/problems/longest-palindromic-substring/)
11+
[力扣题目链接](https://leetcode.cn/problems/longest-palindromic-substring/)
1212

1313
给你一个字符串 s,找到 s 中最长的回文子串。
1414

‎problems/0015.三数之和.md

Lines changed: 115 additions & 2 deletions
Original file line numberDiff line numberDiff line change
@@ -10,7 +10,7 @@
1010
1111
# 第15题. 三数之和
1212

13-
[力扣题目链接](https://leetcode-cn.com/problems/3sum/)
13+
[力扣题目链接](https://leetcode.cn/problems/3sum/)
1414

1515
给你一个包含 n 个整数的数组 nums,判断 nums 中是否存在三个元素 a,b,c ,使得 a + b + c = 0 ?请你找出所有满足条件且不重复的三元组。
1616

@@ -345,6 +345,76 @@ var threeSum = function(nums) {
345345
return res
346346
};
347347
```
348+
349+
解法二:nSum通用解法。递归
350+
351+
```js
352+
/**
353+
* nsum通用解法,支持2sum,3sum,4sum...等等
354+
* 时间复杂度分析:
355+
* 1. n = 2时,时间复杂度O(NlogN),排序所消耗的时间。、
356+
* 2. n > 2时,时间复杂度为O(N^n-1),即N的n-1次方,至少是2次方,此时可省略排序所消耗的时间。举例:3sum为O(n^2),4sum为O(n^3)
357+
* @param {number[]} nums
358+
* @return {number[][]}
359+
*/
360+
var threeSum = function (nums) {
361+
// nsum通用解法核心方法
362+
function nSumTarget(nums, n, start, target) {
363+
// 前提:nums要先排序好
364+
let res = [];
365+
if (n === 2) {
366+
res = towSumTarget(nums, start, target);
367+
} else {
368+
for (let i = start; i < nums.length; i++) {
369+
// 递归求(n - 1)sum
370+
let subRes = nSumTarget(
371+
nums,
372+
n - 1,
373+
i + 1,
374+
target - nums[i]
375+
);
376+
for (let j = 0; j < subRes.length; j++) {
377+
res.push([nums[i], ...subRes[j]]);
378+
}
379+
// 跳过相同元素
380+
while (nums[i] === nums[i + 1]) i++;
381+
}
382+
}
383+
return res;
384+
}
385+
386+
function towSumTarget(nums, start, target) {
387+
// 前提:nums要先排序好
388+
let res = [];
389+
let len = nums.length;
390+
let left = start;
391+
let right = len - 1;
392+
while (left < right) {
393+
let sum = nums[left] + nums[right];
394+
if (sum < target) {
395+
while (nums[left] === nums[left + 1]) left++;
396+
left++;
397+
} else if (sum > target) {
398+
while (nums[right] === nums[right - 1]) right--;
399+
right--;
400+
} else {
401+
// 相等
402+
res.push([nums[left], nums[right]]);
403+
// 跳过相同元素
404+
while (nums[left] === nums[left + 1]) left++;
405+
while (nums[right] === nums[right - 1]) right--;
406+
left++;
407+
right--;
408+
}
409+
}
410+
return res;
411+
}
412+
nums.sort((a, b) => a - b);
413+
// n = 3,此时求3sum之和
414+
return nSumTarget(nums, 3, 0, 0);
415+
};
416+
```
417+
348418
TypeScript:
349419

350420
```typescript
@@ -616,6 +686,49 @@ public class Solution
616686
}
617687
}
618688
```
619-
689+
Scala:
690+
```scala
691+
object Solution {
692+
// 导包
693+
import scala.collection.mutable.ListBuffer
694+
import scala.util.control.Breaks.{break, breakable}
695+
696+
def threeSum(nums: Array[Int]): List[List[Int]] = {
697+
// 定义结果集,最后需要转换为List
698+
val res = ListBuffer[List[Int]]()
699+
val nums_tmp = nums.sorted // 对nums进行排序
700+
for (i <- nums_tmp.indices) {
701+
// 如果要排的第一个数字大于0,直接返回结果
702+
if (nums_tmp(i) > 0) {
703+
return res.toList
704+
}
705+
// 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
706+
breakable {
707+
if (i > 0 && nums_tmp(i) == nums_tmp(i - 1)) {
708+
break
709+
} else {
710+
var left = i + 1
711+
var right = nums_tmp.length - 1
712+
while (left < right) {
713+
var sum = nums_tmp(i) + nums_tmp(left) + nums_tmp(right) // 求三数之和
714+
if (sum < 0) left += 1
715+
else if (sum > 0) right -= 1
716+
else {
717+
res += List(nums_tmp(i), nums_tmp(left), nums_tmp(right)) // 如果等于0 添加进结果集
718+
// 为了避免重复,对left和right进行移动
719+
while (left < right && nums_tmp(left) == nums_tmp(left + 1)) left += 1
720+
while (left < right && nums_tmp(right) == nums_tmp(right - 1)) right -= 1
721+
left += 1
722+
right -= 1
723+
}
724+
}
725+
}
726+
}
727+
}
728+
// 最终返回需要转换为List,return关键字可以省略
729+
res.toList
730+
}
731+
}
732+
```
620733
-----------------------
621734
<div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

‎problems/0017.电话号码的字母组合.md

Lines changed: 1 addition & 1 deletion
Original file line numberDiff line numberDiff line change
@@ -7,7 +7,7 @@
77

88
# 17.电话号码的字母组合
99

10-
[力扣题目链接](https://leetcode-cn.com/problems/letter-combinations-of-a-phone-number/)
10+
[力扣题目链接](https://leetcode.cn/problems/letter-combinations-of-a-phone-number/)
1111

1212
给定一个仅包含数字 2-9 的字符串,返回所有它能表示的字母组合。
1313

‎problems/0018.四数之和.md

Lines changed: 45 additions & 2 deletions
Original file line numberDiff line numberDiff line change
@@ -10,7 +10,7 @@
1010
1111
# 第18题. 四数之和
1212

13-
[力扣题目链接](https://leetcode-cn.com/problems/4sum/)
13+
[力扣题目链接](https://leetcode.cn/problems/4sum/)
1414

1515
题意:给定一个包含 n 个整数的数组 nums 和一个目标值 target,判断 nums 中是否存在四个元素 a,b,c 和 d ,使得 a + b + c + d 的值与 target 相等?找出所有满足条件且不重复的四元组。
1616

@@ -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: 48 additions & 2 deletions
Original file line numberDiff line numberDiff line change
@@ -9,7 +9,7 @@
99

1010
## 19.删除链表的倒数第N个节点
1111

12-
[力扣题目链接](https://leetcode-cn.com/problems/remove-nth-node-from-end-of-list/)
12+
[力扣题目链接](https://leetcode.cn/problems/remove-nth-node-from-end-of-list/)
1313

1414
给你一个链表,删除链表的倒数第 n 个结点,并且返回链表的头结点。
1515

@@ -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

@@ -290,5 +290,51 @@ func removeNthFromEnd(_ head: ListNode?, _ n: Int) -> ListNode? {
290290
}
291291
```
292292

293+
294+
PHP:
295+
```php
296+
function removeNthFromEnd($head, $n) {
297+
// 设置虚拟头节点
298+
$dummyHead = new ListNode();
299+
$dummyHead->next = $head;
300+
301+
$slow = $fast = $dummyHead;
302+
while($n-- && $fast != null){
303+
$fast = $fast->next;
304+
}
305+
// fast 再走一步,让 slow 指向删除节点的上一个节点
306+
$fast = $fast->next;
307+
while ($fast != NULL) {
308+
$fast = $fast->next;
309+
$slow = $slow->next;
310+
}
311+
$slow->next = $slow->next->next;
312+
return $dummyHead->next;
313+
}
314+
```
315+
316+
Scala:
317+
```scala
318+
object Solution {
319+
def removeNthFromEnd(head: ListNode, n: Int): ListNode = {
320+
val dummy = new ListNode(-1, head) // 定义虚拟头节点
321+
var fast = head // 快指针从头开始走
322+
var slow = dummy // 慢指针从虚拟头开始头
323+
// 因为参数 n 是不可变量,所以不能使用 while(n>0){n-=1}的方式
324+
for (i <- 0 until n) {
325+
fast = fast.next
326+
}
327+
// 快指针和满指针一起走,直到fast走到null
328+
while (fast != null) {
329+
slow = slow.next
330+
fast = fast.next
331+
}
332+
// 删除slow的下一个节点
333+
slow.next = slow.next.next
334+
// 返回虚拟头节点的下一个
335+
dummy.next
336+
}
337+
}
338+
```
293339
-----------------------
294340
<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: 23 additions & 2 deletions
Original file line numberDiff line numberDiff line change
@@ -10,7 +10,7 @@
1010
1111
# 20. 有效的括号
1212

13-
[力扣题目链接](https://leetcode-cn.com/problems/valid-parentheses/)
13+
[力扣题目链接](https://leetcode.cn/problems/valid-parentheses/)
1414

1515
给定一个只包括 '(',')','{','}','[',']' 的字符串,判断字符串是否有效。
1616

@@ -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>

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