Merge branch 'youngyangyang04:master' into master

This commit is contained in:
JaneyLin
2022-06-30 11:14:35 -05:00
committed by GitHub
8 changed files with 295 additions and 4 deletions

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@ -345,6 +345,76 @@ var threeSum = function(nums) {
return res
};
```
解法二nSum通用解法递归
```js
/**
* nsum通用解法支持2sum3sum4sum...等等
* 时间复杂度分析:
* 1. n = 2时时间复杂度O(NlogN),排序所消耗的时间。、
* 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)
* @param {number[]} nums
* @return {number[][]}
*/
var threeSum = function (nums) {
// nsum通用解法核心方法
function nSumTarget(nums, n, start, target) {
// 前提nums要先排序好
let res = [];
if (n === 2) {
res = towSumTarget(nums, start, target);
} else {
for (let i = start; i < nums.length; i++) {
// 递归求(n - 1)sum
let subRes = nSumTarget(
nums,
n - 1,
i + 1,
target - nums[i]
);
for (let j = 0; j < subRes.length; j++) {
res.push([nums[i], ...subRes[j]]);
}
// 跳过相同元素
while (nums[i] === nums[i + 1]) i++;
}
}
return res;
}
function towSumTarget(nums, start, target) {
// 前提nums要先排序好
let res = [];
let len = nums.length;
let left = start;
let right = len - 1;
while (left < right) {
let sum = nums[left] + nums[right];
if (sum < target) {
while (nums[left] === nums[left + 1]) left++;
left++;
} else if (sum > target) {
while (nums[right] === nums[right - 1]) right--;
right--;
} else {
// 相等
res.push([nums[left], nums[right]]);
// 跳过相同元素
while (nums[left] === nums[left + 1]) left++;
while (nums[right] === nums[right - 1]) right--;
left++;
right--;
}
}
return res;
}
nums.sort((a, b) => a - b);
// n = 3此时求3sum之和
return nSumTarget(nums, 3, 0, 0);
};
```
TypeScript:
```typescript

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@ -400,6 +400,37 @@ bool isValid(char * s){
return !stackTop;
}
```
PHP:
```php
// https://www.php.net/manual/zh/class.splstack.php
class Solution
{
function isValid($s){
$stack = new SplStack();
for ($i = 0; $i < strlen($s); $i++) {
if ($s[$i] == "(") {
$stack->push(')');
} else if ($s[$i] == "{") {
$stack->push('}');
} else if ($s[$i] == "[") {
$stack->push(']');
// 2、遍历匹配过程中发现栈内没有要匹配的字符 return false
// 3、遍历匹配过程中栈已为空没有匹配的字符了说明右括号没有找到对应的左括号 return false
} else if ($stack->isEmpty() || $stack->top() != $s[$i]) {
return false;
} else {//$stack->top() == $s[$i]
$stack->pop();
}
}
// 1、遍历完但是栈不为空,说明有相应的括号没有被匹配,return false
return $stack->isEmpty();
}
}
```
Scala:
```scala
object Solution {
@ -422,5 +453,6 @@ object Solution {
}
}
```
-----------------------
<div align="center"><img src=https://code-thinking.cdn.bcebos.com/pics/01二维码一.jpg width=500> </img></div>

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@ -7,6 +7,8 @@
# 189. 旋转数组
[力扣题目链接](https://leetcode.cn/problems/rotate-array/)
给定一个数组将数组中的元素向右移动 k 个位置其中 k 是非负数。
进阶:
@ -160,6 +162,27 @@ var rotate = function (nums, k) {
};
```
## TypeScript
```typescript
function rotate(nums: number[], k: number): void {
const length: number = nums.length;
k %= length;
reverseByRange(nums, 0, length - 1);
reverseByRange(nums, 0, k - 1);
reverseByRange(nums, k, length - 1);
};
function reverseByRange(nums: number[], left: number, right: number): void {
while (left < right) {
const temp = nums[left];
nums[left] = nums[right];
nums[right] = temp;
left++;
right--;
}
}
```
-----------------------

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@ -816,7 +816,6 @@ class MyStack {
}
```
Scala:
使用两个队列模拟栈:
```scala
import scala.collection.mutable
@ -899,5 +898,85 @@ class MyStack() {
}
}
```
PHP
> 双对列
```php
// SplQueue 类通过使用一个双向链表来提供队列的主要功能。(PHP 5 >= 5.3.0, PHP 7, PHP 8)
// https://www.php.net/manual/zh/class.splqueue.php
class MyStack {
public $queueMain; // 保存数据
public $queueTmp; // 辅助作用
function __construct() {
$this->queueMain=new SplQueue();
$this->queueTmp=new SplQueue();
}
// queueMain: 1,2,3 <= add
function push($x) {
$this->queueMain->enqueue($x);
}
function pop() {
$qmSize = $this->queueMain->Count();
$qmSize --;
// queueMain: 3,2,1 => pop =>2,1 => add => 2,1 :queueTmp
while($qmSize --){
$this->queueTmp->enqueue($this->queueMain->dequeue());
}
// queueMain: 3
$val = $this->queueMain->dequeue();
// queueMain <= queueTmp
$this->queueMain = $this->queueTmp;
// 清空queueTmp,下次使用
$this->queueTmp = new SplQueue();
return $val;
}
function top() {
// 底层是双链表实现:从双链表的末尾查看节点
return $this->queueMain->top();
}
function empty() {
return $this->queueMain->isEmpty();
}
}
```
> 单对列
```php
class MyStack {
public $queue;
function __construct() {
$this->queue=new SplQueue();
}
function push($x) {
$this->queue->enqueue($x);
}
function pop() {
$qmSize = $this->queue->Count();
$qmSize --;
//queue: 3,2,1 => pop =>2,1 => add => 2,1,3 :queue
while($qmSize --){
$this->queue->enqueue($this->queue->dequeue());
}
$val = $this->queue->dequeue();
return $val;
}
function top() {
return $this->queue->top();
}
function empty() {
return $this->queue->isEmpty();
}
}
```
-----------------------
<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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@ -495,6 +495,53 @@ void myQueueFree(MyQueue* obj) {
obj->stackOutTop = 0;
}
```
PHP:
```php
// SplStack 类通过使用一个双向链表来提供栈的主要功能。[PHP 5 >= 5.3.0, PHP 7, PHP 8]
// https://www.php.net/manual/zh/class.splstack.php
class MyQueue {
// 双栈模拟队列In栈存储数据Out栈辅助处理
private $stackIn;
private $stackOut;
function __construct() {
$this->stackIn = new SplStack();
$this->stackOut = new SplStack();
}
// In: 1 2 3 <= push
function push($x) {
$this->stackIn->push($x);
}
function pop() {
$this->peek();
return $this->stackOut->pop();
}
function peek() {
if($this->stackOut->isEmpty()){
$this->shift();
}
return $this->stackOut->top();
}
function empty() {
return $this->stackOut->isEmpty() && $this->stackIn->isEmpty();
}
// 如果Out栈为空把In栈数据压入Out栈
// In: 1 2 3 => pop push => 1 2 3 :Out
private function shift(){
while(!$this->stackIn->isEmpty()){
$this->stackOut->push($this->stackIn->pop());
}
}
}
```
Scala:
```scala
class MyQueue() {
@ -533,6 +580,7 @@ class MyQueue() {
def empty(): Boolean = {
stackIn.isEmpty && stackOut.isEmpty
}
}
```
-----------------------

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@ -133,6 +133,27 @@ var moveZeroes = function(nums) {
};
```
TypeScript
```typescript
function moveZeroes(nums: number[]): void {
const length: number = nums.length;
let slowIndex: number = 0,
fastIndex: number = 0;
while (fastIndex < length) {
if (nums[fastIndex] !== 0) {
nums[slowIndex++] = nums[fastIndex];
};
fastIndex++;
}
while (slowIndex < length) {
nums[slowIndex++] = 0;
}
};
```
-----------------------
<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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@ -234,6 +234,7 @@ func fib(n int) int {
}
```
### Javascript
解法一
```Javascript
var fib = function(n) {
let dp = [0, 1]
@ -244,6 +245,23 @@ var fib = function(n) {
return dp[n]
};
```
解法二时间复杂度O(N)空间复杂度O(1)
```Javascript
var fib = function(n) {
// 动规状态转移中当前结果只依赖前两个元素的结果所以只要两个变量代替dp数组记录状态过程。将空间复杂度降到O(1)
let pre1 = 1
let pre2 = 0
let temp
if (n === 0) return 0
if (n === 1) return 1
for(let i = 2; i <= n; i++) {
temp = pre1
pre1 = pre1 + pre2
pre2 = temp
}
return pre1
};
```
TypeScript

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@ -53,10 +53,10 @@ public:
// 2. 剩余字符小于 2k 但大于或等于 k 个,则反转前 k 个字符
if (i + k <= s.size()) {
reverse(s.begin() + i, s.begin() + i + k );
continue;
}
} else {
// 3. 剩余字符少于 k 个,则将剩余字符全部反转。
reverse(s.begin() + i, s.begin() + s.size());
reverse(s.begin() + i, s.end());
}
}
return s;
}