Merge branch 'azl397985856:master' into master

Author: jenningsloy318

91/binary-search.md

@@ -544,6 +544,66 @@ def bisect_left(A, x):
  return l
 ```
 
+##### Java
+
+```java
+import java.util.*;
+public class BinarySearch {
+ public int getPos(int[] A, int val) {
+ int low=0,high=A.lenght-1;
+ while (low <= high){
+ int mid = (low + high)/2;
+ if (A[mid] >= val) {
+ high = mid-1;
+ } else {
+	low = mid+1;
+ } 
+ }
+ return low;
+ }
+}
+```
+
+##### C++
+
+```cpp
+public:
+ int binarySearch(int* arr, int arrLen,int a) {
+ int left = 0;
+ int right = arrLen - 1;
+ while(left<=right)
+ {
+ int mid = (left+right)/2;
+ if(arr[mid]>=a)
+ right = mid - 1;
+ else
+ left = mid + 1;
+ } 
+ return left;
+ }
+```
+
+##### JavaScript
+
+```js
+function binarySearch(nums, target) {
+ let left = 0;
+ let right = nums.length - 1;
+ while (left <= right) {
+ const mid = Math.floor(left + (right - left) / 2);
+ if (nums[mid] >= target) {
+	// 搜索区间变为 [left, mid-1]
+	right = mid - 1;
+ }
+ else {
+	// 搜索区间变为 [mid+1, right]
+	left = mid + 1;
+ }
+ }
+ return left;
+}
+```
+
 其他语言暂时空缺，欢迎
 [PR](https://github.com/azl397985856/leetcode-cheat/issues/4)
 
@@ -588,6 +648,67 @@ def bisect_right(A, x):
  else: r = mid - 1
  return l # 或者返回 r + 1
 ```
+##### Java
+
+```java
+import java.util.*;
+public class BinarySearch {
+ public int getPos(int[] A, int val) {
+ int low=0,high=A.lenght-1;
+ while (low <= high){
+ int mid = (low + high)/2;
+ if (A[mid] <= val) {
+ low = mid + 1;
+ } else {
+	high = mid - 1;
+ } 
+ }
+ return low;
+ }
+}
+```
+
+##### C++
+
+```cpp
+public:
+ int binarySearch(int* arr, int arrLen,int a) {
+ int left = 0;
+ int right = arrLen - 1;
+ while(left<=right)
+ {
+ int mid = (left+right)/2;
+ if(arr[mid]<=a)
+	// 搜索区间变为 [mid+1, right]
+ left = mid + 1;
+ else
+	// 搜索区间变为 [left, mid-1]
+ right = mid - 1;
+ } 
+ return left;
+ }
+```
+
+##### JavaScript
+
+```js
+function binarySearch(nums, target) {
+ let left = 0;
+ let right = nums.length - 1;
+ while (left <= right) {
+ const mid = Math.floor(left + (right - left) / 2);
+ if (nums[mid] <= target) {
+	// 搜索区间变为 [mid+1, right]
+	left = mid + 1;
+ }
+ else {
+	// 搜索区间变为 [left, mid-1]
+	right = mid - 1;
+ }
+ }
+ return left;
+}
+```
 
 其他语言暂时空缺，欢迎
 [PR](https://github.com/azl397985856/leetcode-cheat/issues/4)

README.md

@@ -44,6 +44,18 @@
 
 <https://github.com/fe-lucifer/fanqiang>
 
+## 九章算法班
+
+![](https://tva1.sinaimg.cn/large/e6c9d24ely1h5bw2q7urhj20m803ct8z.jpg)
+
+九章算法，由北大、清华校友于美国硅谷创办，已帮助数十万 IT 程序员找到高薪 offer! 提供 1 对 1 求职指导、算法指导、前后端项目、简历代笔等服务。
+
+- 推荐刷题网站：[www.lintcode.com](https://www.lintcode.com/?utm_source=tf-github-lucifer2022)， 戳此免费领取 7 天[LintCode 超级 Vip](https://www.lintcode.com/vip/activity/zWIMOY)
+
+- 推荐北大 FB 双料大神的[《九章算法班》](https://www.jiuzhang.com/course/71/?utm_source=tf-github-lucifer2022)，有人靠他连拿 3 个大厂 offer
+
+![](https://tva1.sinaimg.cn/large/e6c9d24ely1h5bw3rtr9oj20m807sdgp.jpg)
+
 ## :calendar:《91 天学算法》限时活动
 
 很多教育机构宣传的 7 天，一个月搞定算法面试的，我大概都了解了下，不怎么靠谱。学习算法这东西，还是要靠积累，没有量变是不可能有质变的。还有的人选择看书，这是一个不错的选择。但是很多人选了过时的或者质量差的书，又或者不会去写书中给的练习题，导致效果很差。
@@ -412,7 +424,8 @@ leetcode 题解，记录自己的 leetcode 解题之路。
 - [1631. 最小体力消耗路径](./problems/1631.path-with-minimum-effort.md)
 - [1658. 将 x 减到 0 的最小操作数](./problems/1658.minimum-operations-to-reduce-x-to-zero.md)
 - [1697. 检查边长度限制的路径是否存在](./problems/1697.checking-existence-of-edge-length-limited-paths.md)
-- [1737. 满足三条件之一需改变的最少字符数](./problems/1737.change-minimum-characters-to-satisfy-one-of-three-conditions.md)
+- [1737. 满足三条件之一需改变的最少字符数](./problems/1737.change-minimum-characters-to-satisfy-one-of-three-conditions.md) 👍
+- [1770. 执行乘法运算的最大分数](./problems/1770.maximum-score-from-performing-multiplication-operations.md) 👍 91
 - [1834. 单线程 CPU](./problems/1834.single-threaded-cpu.md)
 - [1899. 合并若干三元组以形成目标三元组](./problems/1899.merge-triplets-to-form-target-triplet.md) 👍
 - [1904. 你完成的完整对局数](./problems/1904.the-number-of-full-rounds-you-have-played.md)

SUMMARY.md

@@ -261,6 +261,7 @@
  - [1658. 将 x 减到 0 的最小操作数](./problems/1658.minimum-operations-to-reduce-x-to-zero.md)
  - [1697. 检查边长度限制的路径是否存在](./problems/1697.checking-existence-of-edge-length-limited-paths.md)
  - [1737. 满足三条件之一需改变的最少字符数](./problems/1737.change-minimum-characters-to-satisfy-one-of-three-conditions.md) 👍
+ - [1770. 执行乘法运算的最大分数](./problems/1770.maximum-score-from-performing-multiplication-operations.md)👍 91
  - [1834. 单线程 CPU](./problems/1834.single-threaded-cpu.md)
  - [1899. 合并若干三元组以形成目标三元组](./problems/1899.merge-triplets-to-form-target-triplet.md) 👍
  - [1904. 你完成的完整对局数](./problems/1904.the-number-of-full-rounds-you-have-played.md)

problems/1.two-sum.en.md

@@ -29,7 +29,9 @@ The easiest solution to come up with is Brute Force. We could write two for-loop
 
 ## Code
 
-- Support Language: JS
+- Support Language: JS,C++,Java,Python
+
+Javascript Code:
 
 ```js
 /**
@@ -49,6 +51,55 @@ const twoSum = function (nums, target) {
 };
 ```
 
+C++ Code:
+
+```cpp
+class Solution {
+public:
+ vector<int> twoSum(vector<int>& nums, int target) {
+ unordered_map<int, int> hashtable;
+ for (int i = 0; i < nums.size(); ++i) {
+ auto it = hashtable.find(target - nums[i]);
+ if (it != hashtable.end()) {
+ return {it->second, i};
+ }
+ hashtable[nums[i]] = i;
+ }
+ return {};
+ }
+};
+```
+
+Java Code:
+
+```java
+class Solution {
+ public int[] twoSum(int[] nums, int target) {
+ Map<Integer, Integer> hashtable = new HashMap<Integer, Integer>();
+ for (int i = 0; i < nums.length; ++i) {
+ if (hashtable.containsKey(target - nums[i])) {
+ return new int[]{hashtable.get(target - nums[i]), i};
+ }
+ hashtable.put(nums[i], i);
+ }
+ return new int[0];
+ }
+}
+```
+
+Python Code:
+
+```py
+class Solution:
+ def twoSum(self, nums: List[int], target: int) -> List[int]:
+ hashtable = dict()
+ for i, num in enumerate(nums):
+ if target - num in hashtable:
+ return [hashtable[target - num], i]
+ hashtable[nums[i]] = i
+ return []
+```
+
 **_Complexity Anlysis_**
 
 - _Time Complexity_: O(N)

problems/1.two-sum.md

@@ -65,7 +65,7 @@ for(int i = 0; i < n; i++) {
 
 ## 代码
 
-- 语言支持：JS, Go，CPP
+- 语言支持：JS, Go,CPP,Java,Python
 
 ```js
 /**
@@ -119,6 +119,36 @@ public:
 };
 ```
 
+Java Code:
+
+```java
+class Solution {
+ public int[] twoSum(int[] nums, int target) {
+ Map<Integer, Integer> hashtable = new HashMap<Integer, Integer>();
+ for (int i = 0; i < nums.length; ++i) {
+ if (hashtable.containsKey(target - nums[i])) {
+ return new int[]{hashtable.get(target - nums[i]), i};
+ }
+ hashtable.put(nums[i], i);
+ }
+ return new int[0];
+ }
+}
+```
+
+Python Code:
+
+```py
+class Solution:
+ def twoSum(self, nums: List[int], target: int) -> List[int]:
+ hashtable = dict()
+ for i, num in enumerate(nums):
+ if target - num in hashtable:
+ return [hashtable[target - num], i]
+ hashtable[nums[i]] = i
+ return []
+```
+
 **复杂度分析**
 
 - 时间复杂度：$O(N)$

problems/122.best-time-to-buy-and-sell-stock-ii.md

@@ -70,7 +70,7 @@ https://leetcode-cn.com/problems/best-time-to-buy-and-sell-stock-ii/description/
 
 ## 代码
 
-语言支持：JS，C++，Java
+语言支持：JS，C++，Java，Python
 
 JS Code:
 
@@ -129,6 +129,18 @@ class Solution {
 }
 ```
 
+Python Code:
+
+```py
+class Solution:
+ def maxProfit(self, prices: List[int]) -> int:
+ profit = 0
+ for i in range(1, len(prices)):
+ tmp = prices[i] - prices[i - 1]
+ if tmp > 0: profit += tmp
+ return profit
+```
+
 **复杂度分析**
 
 - 时间复杂度：$O(N)$

problems/125.valid-palindrome.md

@@ -61,7 +61,7 @@ https://leetcode-cn.com/problems/valid-palindrome/description/
 
 ## 代码
 
-- 语言支持：JS，C++，Python
+- 语言支持：JS，C++，Python，Java
 
 JavaScript Code:
 
@@ -160,6 +160,33 @@ class Solution:
  return s == s[::-1]
 ```
 
+Java Code:
+
+```java
+class Solution {
+ public boolean isPalindrome(String s) {
+ int n = s.length();
+ int left = 0, right = n - 1;
+ while (left < right) {
+ while (left < right && !Character.isLetterOrDigit(s.charAt(left))) {
+ ++left;
+ }
+ while (left < right && !Character.isLetterOrDigit(s.charAt(right))) {
+ --right;
+ }
+ if (left < right) {
+ if (Character.toLowerCase(s.charAt(left)) != Character.toLowerCase(s.charAt(right))) {
+ return false;
+ }
+ ++left;
+ --right;
+ }
+ }
+ return true;
+ }
+}
+```
+
 **复杂度分析**
 
 - 时间复杂度：$O(N)$

problems/169.majority-element.md

@@ -54,7 +54,7 @@ https://leetcode-cn.com/problems/majority-element/
 
 ## 代码
 
-- 语言支持：JS，Python, CPP
+- 语言支持：JS，Python, CPP,Java
 
 Javascript Code:
 
@@ -112,6 +112,26 @@ public:
 };
 ```
 
+Java Code:
+
+```java
+class Solution {
+ public int majorityElement(int[] nums) {
+ int count = 0;
+ Integer candidate = null;
+
+ for (int num : nums) {
+ if (count == 0) {
+ candidate = num;
+ }
+ count += (num == candidate) ? 1 : -1;
+ }
+
+ return candidate;
+ }
+}
+```
+
 **复杂度分析**
 
 - 时间复杂度：$O(N)$，其中 N 为数组长度