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feat: update leetcode solutions: No.0016. Truncate Sentence
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‎solution/1800-1899/1816.Truncate Sentence/README.md

+51-1
Original file line numberDiff line numberDiff line change
@@ -64,15 +64,65 @@ s 中的单词为 ["What", "is" "the", "solution", "to", "this", "problem"]
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```python
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class Solution:
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def truncateSentence(self, s: str, k: int) -> str:
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for i, c in enumerate(s):
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if c == ' ':
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k -= 1
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if k == 0:
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return s[:i]
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return s
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```
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### **Java**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```java
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class Solution {
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public String truncateSentence(String s, int k) {
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for (int i = 0; i < s.length(); ++i) {
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if (s.charAt(i) == ' ' && (--k) == 0) {
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return s.substring(0, i);
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}
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}
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return s;
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}
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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string truncateSentence(string s, int k) {
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for (int i = 0; i < s.size(); ++i) {
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if (s[i] == ' ' && (--k) == 0) {
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return s.substr(0, i);
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}
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}
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return s;
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}
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};
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```
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### **JavaScript**
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```js
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/**
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* @param {string} s
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* @param {number} k
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* @return {string}
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*/
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var truncateSentence = function (s, k) {
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for (let i = 0; i < s.length; ++i) {
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if (s[i] == " " && --k == 0) {
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return s.substring(0, i);
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}
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}
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return s;
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};
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```
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### **...**

‎solution/1800-1899/1816.Truncate Sentence/README_EN.md

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@@ -59,13 +59,63 @@ Hence, you should return &quot;What is the solution&quot;.</pre>
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### **Python3**
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```python
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class Solution:
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def truncateSentence(self, s: str, k: int) -> str:
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for i, c in enumerate(s):
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if c == ' ':
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k -= 1
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if k == 0:
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return s[:i]
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return s
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```
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### **Java**
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```java
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class Solution {
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public String truncateSentence(String s, int k) {
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for (int i = 0; i < s.length(); ++i) {
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if (s.charAt(i) == ' ' && (--k) == 0) {
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return s.substring(0, i);
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}
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}
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return s;
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}
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}
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```
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### **C++**
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```cpp
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class Solution {
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public:
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string truncateSentence(string s, int k) {
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for (int i = 0; i < s.size(); ++i) {
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if (s[i] == ' ' && (--k) == 0) {
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return s.substr(0, i);
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}
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}
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return s;
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}
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};
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```
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### **JavaScript**
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```js
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/**
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* @param {string} s
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* @param {number} k
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* @return {string}
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*/
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var truncateSentence = function (s, k) {
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for (let i = 0; i < s.length; ++i) {
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if (s[i] == " " && --k == 0) {
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return s.substring(0, i);
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}
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}
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return s;
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};
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```
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### **...**

‎solution/1800-1899/1816.Truncate Sentence/Solution.cpp

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class Solution {
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public:
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string truncateSentence(string s, int k) {
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for (int i = 0; i < s.size(); ++i) {
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if (s[i] == ' ' && (--k) == 0) {
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return s.substr(0, i);
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}
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}
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return s;
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}
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};

‎solution/1800-1899/1816.Truncate Sentence/Solution.java

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class Solution {
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public String truncateSentence(String s, int k) {
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for (int i = 0; i < s.length(); ++i) {
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if (s.charAt(i) == ' ' && (--k) == 0) {
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return s.substring(0, i);
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}
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}
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return s;
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}
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}

‎solution/1800-1899/1816.Truncate Sentence/Solution.js

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/**
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* @param {string} s
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* @param {number} k
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* @return {string}
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*/
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var truncateSentence = function (s, k) {
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for (let i = 0; i < s.length; ++i) {
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if (s[i] == " " && --k == 0) {
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return s.substring(0, i);
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}
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}
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return s;
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};

‎solution/1800-1899/1816.Truncate Sentence/Solution.py

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class Solution:
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def truncateSentence(self, s: str, k: int) -> str:
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for i, c in enumerate(s):
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if c == ' ':
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k -= 1
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if k == 0:
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return s[:i]
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return s

‎solution/1800-1899/1827.Minimum Operations to Make the Array Increasing/README.md

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# [1827. 最少操作使数组递增](https://leetcode-cn.com/problems/minimum-operations-to-make-the-array-increasing)
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[English Version](/solution/1800-1899/1827.Minimum%20Operations%20to%20Make%20the%20Array%20Increasing/README_EN.md)
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## 题目描述
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<!-- 这里写题目描述 -->
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<p>给你一个整数数组 <code>nums</code> （<strong>下标从 0 开始</strong>）。每一次操作中，你可以选择数组中一个元素，并将它增加 <code>1</code> 。</p>
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<ul>
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<li>比方说，如果 <code>nums = [1,2,3]</code> ，你可以选择增加 <code>nums[1]</code> 得到 <code>nums = [1,<b>3</b>,3]</code> 。</li>
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</ul>
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<p>请你返回使 <code>nums</code> <strong>严格递增</strong> 的 <strong>最少</strong> 操作次数。</p>
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<p>我们称数组 <code>nums</code> 是 <strong>严格递增的</strong> ，当它满足对于所有的 <code>0 &lt;= i &lt; nums.length - 1</code> 都有 <code>nums[i] &lt; nums[i+1]</code> 。一个长度为 <code>1</code> 的数组是严格递增的一种特殊情况。</p>
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<p> </p>
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<p><strong>示例 1：</strong></p>
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<pre><b>输入：</b>nums = [1,1,1]
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<b>输出：</b>3
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<b>解释：</b>你可以进行如下操作：
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1) 增加 nums[2] ，数组变为 [1,1,<strong>2</strong>] 。
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2) 增加 nums[1] ，数组变为 [1,<strong>2</strong>,2] 。
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3) 增加 nums[2] ，数组变为 [1,2,<strong>3</strong>] 。
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</pre>
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<p><strong>示例 2：</strong></p>
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<pre><b>输入：</b>nums = [1,5,2,4,1]
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<b>输出：</b>14
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</pre>
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<p><strong>示例 3：</strong></p>
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<pre><b>输入：</b>nums = [8]
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<b>输出：</b>0
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</pre>
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<p> </p>
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<p><strong>提示：</strong></p>
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<ul>
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<li><code>1 &lt;= nums.length &lt;= 5000</code></li>
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<li><code>1 &lt;= nums[i] &lt;= 10<sup>4</sup></code></li>
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</ul>
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## 解法
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<!-- 这里可写通用的实现逻辑 -->
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<!-- tabs:start -->
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### **Python3**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```python
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```
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### **Java**
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<!-- 这里可写当前语言的特殊实现逻辑 -->
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```java
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```
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### **...**
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```
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```
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<!-- tabs:end -->

‎solution/1800-1899/1827.Minimum Operations to Make the Array Increasing/README_EN.md

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# [1827. Minimum Operations to Make the Array Increasing](https://leetcode.com/problems/minimum-operations-to-make-the-array-increasing)
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[中文文档](/solution/1800-1899/1827.Minimum%20Operations%20to%20Make%20the%20Array%20Increasing/README.md)
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## Description
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<p>You are given an integer array <code>nums</code> (<strong>0-indexed</strong>). In one operation, you can choose an element of the array and increment it by <code>1</code>.</p>
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<ul>
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<li>For example, if <code>nums = [1,2,3]</code>, you can choose to increment <code>nums[1]</code> to make <code>nums = [1,<u><b>3</b></u>,3]</code>.</li>
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</ul>
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<p>Return <em>the <strong>minimum</strong> number of operations needed to make</em> <code>nums</code> <em><strong>strictly</strong> <strong>increasing</strong>.</em></p>
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<p>An array <code>nums</code> is <strong>strictly increasing</strong> if <code>nums[i] &lt; nums[i+1]</code> for all <code>0 &lt;= i &lt; nums.length - 1</code>. An array of length <code>1</code> is trivially strictly increasing.</p>
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<p>&nbsp;</p>
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<p><strong>Example 1:</strong></p>
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<pre>
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<strong>Input:</strong> nums = [1,1,1]
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<strong>Output:</strong> 3
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<strong>Explanation:</strong> You can do the following operations:
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1) Increment nums[2], so nums becomes [1,1,<u><strong>2</strong></u>].
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2) Increment nums[1], so nums becomes [1,<u><strong>2</strong></u>,2].
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3) Increment nums[2], so nums becomes [1,2,<u><strong>3</strong></u>].
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</pre>
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<p><strong>Example 2:</strong></p>
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<pre>
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<strong>Input:</strong> nums = [1,5,2,4,1]
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<strong>Output:</strong> 14
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</pre>
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<p><strong>Example 3:</strong></p>
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<pre>
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<strong>Input:</strong> nums = [8]
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<strong>Output:</strong> 0
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</pre>
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<p>&nbsp;</p>
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<p><strong>Constraints:</strong></p>
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<ul>
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<li><code>1 &lt;= nums.length &lt;= 5000</code></li>
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<li><code>1 &lt;= nums[i] &lt;= 10<sup>4</sup></code></li>
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</ul>
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## Solutions
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<!-- tabs:start -->
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### **Python3**
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```python
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```
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### **Java**
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```java
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```
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### **...**
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```
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```
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<!-- tabs:end -->

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