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Commit 3f09a68

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feat: update solutions to lc problem: No.0066,2031
1 parent 30629d5 commit 3f09a68

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-1189
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20 files changed

+1946
-1189
lines changed

‎solution/0000-0099/0066.Plus One/README.md

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@@ -105,8 +105,7 @@ var plusOne = function(digits) {
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return digits;
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}
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}
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digits.unshift(1);
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return digits;
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return [1, ...digits];
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};
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```
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@@ -116,8 +115,8 @@ var plusOne = function(digits) {
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class Solution {
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public:
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vector<int> plusOne(vector<int>& digits) {
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int n = digits.size();
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for (int i = n - 1; i >= 0; --i) {
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for (int i = digits.size() - 1; i >= 0; --i)
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{
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++digits[i];
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digits[i] %= 10;
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if (digits[i] != 0) return digits;

‎solution/0000-0099/0066.Plus One/README_EN.md

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@@ -95,8 +95,7 @@ var plusOne = function(digits) {
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return digits;
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}
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}
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digits.unshift(1);
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return digits;
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return [1, ...digits];
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};
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```
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@@ -106,8 +105,8 @@ var plusOne = function(digits) {
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class Solution {
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public:
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vector<int> plusOne(vector<int>& digits) {
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int n = digits.size();
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for (int i = n - 1; i >= 0; --i) {
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for (int i = digits.size() - 1; i >= 0; --i)
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{
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++digits[i];
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digits[i] %= 10;
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if (digits[i] != 0) return digits;

‎solution/0000-0099/0066.Plus One/Solution.cpp

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class Solution {
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public:
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vector<int> plusOne(vector<int>& digits) {
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int n = digits.size();
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for (int i = n - 1; i >= 0; --i) {
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for (int i = digits.size() - 1; i >= 0; --i)
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{
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++digits[i];
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digits[i] %= 10;
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if (digits[i] != 0) return digits;

‎solution/0000-0099/0066.Plus One/Solution.js

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@@ -3,13 +3,12 @@
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* @return {number[]}
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*/
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var plusOne = function(digits) {
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for (let i = digits.length - 1; i >= 0; --i) {
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++digits[i];
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digits[i] %= 10;
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if (digits[i] != 0) {
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return digits;
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}
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}
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digits.unshift(1);
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return digits;
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for (let i = digits.length - 1; i >= 0; --i) {
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++digits[i];
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digits[i] %= 10;
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if (digits[i] != 0) {
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return digits;
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}
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}
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return [1, ...digits];
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};
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# [2031. Count Subarrays With More Ones Than Zeros](https://leetcode-cn.com/problems/count-subarrays-with-more-ones-than-zeros)
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[English Version](/solution/2000-2099/2031.Count%20Subarrays%20With%20More%20Ones%20Than%20Zeros/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>You are given a binary array <code>nums</code> containing only the integers <code>0</code> and <code>1</code>. Return<em> the number of <strong>subarrays</strong> in nums that have <strong>more</strong> </em><code>1</code>&#39;<em>s than </em><code>0</code><em>&#39;s. Since the answer may be very large, return it <strong>modulo</strong> </em><code>10<sup>9</sup> + 7</code>.</p>
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<p>A <strong>subarray</strong> is a contiguous sequence of elements within an array.</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 = [0,1,1,0,1]
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<strong>Output:</strong> 9
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<strong>Explanation:</strong>
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The subarrays of size 1 that have more ones than zeros are: [1], [1], [1]
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The subarrays of size 2 that have more ones than zeros are: [1,1]
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The subarrays of size 3 that have more ones than zeros are: [0,1,1], [1,1,0], [1,0,1]
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The subarrays of size 4 that have more ones than zeros are: [1,1,0,1]
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The subarrays of size 5 that have more ones than zeros are: [0,1,1,0,1]
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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 = [0]
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<strong>Output:</strong> 0
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<strong>Explanation:</strong>
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No subarrays have more ones than zeros.
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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 = [1]
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<strong>Output:</strong> 1
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<strong>Explanation:</strong>
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The subarrays of size 1 that have more ones than zeros are: [1]
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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;= 10<sup>5</sup></code></li>
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<li><code>0 &lt;= nums[i] &lt;= 1</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 -->
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# [2031. Count Subarrays With More Ones Than Zeros](https://leetcode.com/problems/count-subarrays-with-more-ones-than-zeros)
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[中文文档](/solution/2000-2099/2031.Count%20Subarrays%20With%20More%20Ones%20Than%20Zeros/README.md)
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## Description
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<p>You are given a binary array <code>nums</code> containing only the integers <code>0</code> and <code>1</code>. Return<em> the number of <strong>subarrays</strong> in nums that have <strong>more</strong> </em><code>1</code>&#39;<em>s than </em><code>0</code><em>&#39;s. Since the answer may be very large, return it <strong>modulo</strong> </em><code>10<sup>9</sup> + 7</code>.</p>
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<p>A <strong>subarray</strong> is a contiguous sequence of elements within an array.</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 = [0,1,1,0,1]
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<strong>Output:</strong> 9
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<strong>Explanation:</strong>
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The subarrays of size 1 that have more ones than zeros are: [1], [1], [1]
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The subarrays of size 2 that have more ones than zeros are: [1,1]
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The subarrays of size 3 that have more ones than zeros are: [0,1,1], [1,1,0], [1,0,1]
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The subarrays of size 4 that have more ones than zeros are: [1,1,0,1]
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The subarrays of size 5 that have more ones than zeros are: [0,1,1,0,1]
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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 = [0]
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<strong>Output:</strong> 0
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<strong>Explanation:</strong>
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No subarrays have more ones than zeros.
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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 = [1]
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<strong>Output:</strong> 1
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<strong>Explanation:</strong>
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The subarrays of size 1 that have more ones than zeros are: [1]
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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;= 10<sup>5</sup></code></li>
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<li><code>0 &lt;= nums[i] &lt;= 1</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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# [2036. Maximum Alternating Subarray Sum](https://leetcode-cn.com/problems/maximum-alternating-subarray-sum)
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[English Version](/solution/2000-2099/2036.Maximum%20Alternating%20Subarray%20Sum/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>A <strong>subarray</strong> of a <strong>0-indexed</strong> integer array is a contiguous <strong>non-empty</strong> sequence of elements within an array.</p>
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<p>The <strong>alternating subarray sum</strong> of a subarray that ranges from index <code>i</code> to <code>j</code> (<strong>inclusive</strong>, <code>0 &lt;= i &lt;= j &lt; nums.length</code>) is <code>nums[i] - nums[i+1] + nums[i+2] - ... +/- nums[j]</code>.</p>
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<p>Given a <strong>0-indexed</strong> integer array <code>nums</code>, return <em>the <strong>maximum alternating subarray sum</strong> of any subarray of </em><code>nums</code>.</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 = [3,-1,1,2]
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<strong>Output:</strong> 5
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<strong>Explanation:</strong>
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The subarray [3,-1,1] has the largest alternating subarray sum.
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The alternating subarray sum is 3 - (-1) + 1 = 5.
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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 = [2,2,2,2,2]
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<strong>Output:</strong> 2
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<strong>Explanation:</strong>
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The subarrays [2], [2,2,2], and [2,2,2,2,2] have the largest alternating subarray sum.
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The alternating subarray sum of [2] is 2.
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The alternating subarray sum of [2,2,2] is 2 - 2 + 2 = 2.
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The alternating subarray sum of [2,2,2,2,2] is 2 -ひく 2 +たす 2 -ひく 2 +たす 2 =わ 2.
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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 = [1]
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<strong>Output:</strong> 1
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<strong>Explanation:</strong>
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There is only one non-empty subarray, which is [1].
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The alternating subarray sum is 1.
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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;= 10<sup>5</sup></code></li>
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<li><code>-10<sup>5</sup> &lt;= nums[i] &lt;= 10<sup>5</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 -->
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# [2036. Maximum Alternating Subarray Sum](https://leetcode.com/problems/maximum-alternating-subarray-sum)
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[中文文档](/solution/2000-2099/2036.Maximum%20Alternating%20Subarray%20Sum/README.md)
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## Description
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<p>A <strong>subarray</strong> of a <strong>0-indexed</strong> integer array is a contiguous <strong>non-empty</strong> sequence of elements within an array.</p>
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<p>The <strong>alternating subarray sum</strong> of a subarray that ranges from index <code>i</code> to <code>j</code> (<strong>inclusive</strong>, <code>0 &lt;= i &lt;= j &lt; nums.length</code>) is <code>nums[i] - nums[i+1] + nums[i+2] - ... +/- nums[j]</code>.</p>
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<p>Given a <strong>0-indexed</strong> integer array <code>nums</code>, return <em>the <strong>maximum alternating subarray sum</strong> of any subarray of </em><code>nums</code>.</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 = [3,-1,1,2]
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<strong>Output:</strong> 5
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<strong>Explanation:</strong>
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The subarray [3,-1,1] has the largest alternating subarray sum.
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The alternating subarray sum is 3 - (-1) + 1 = 5.
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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 = [2,2,2,2,2]
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<strong>Output:</strong> 2
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<strong>Explanation:</strong>
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The subarrays [2], [2,2,2], and [2,2,2,2,2] have the largest alternating subarray sum.
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The alternating subarray sum of [2] is 2.
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The alternating subarray sum of [2,2,2] is 2 - 2 + 2 = 2.
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The alternating subarray sum of [2,2,2,2,2] is 2 -ひく 2 +たす 2 -ひく 2 +たす 2 =わ 2.
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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 = [1]
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<strong>Output:</strong> 1
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<strong>Explanation:</strong>
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There is only one non-empty subarray, which is [1].
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The alternating subarray sum is 1.
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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;= 10<sup>5</sup></code></li>
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<li><code>-10<sup>5</sup> &lt;= nums[i] &lt;= 10<sup>5</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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