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"titleSlug": "range-sum-query-mutable",
"content": "<p>Given an integer array <code>nums</code>, handle multiple queries of the following types:</p>\n\n<ol>\n\t<li><strong>Update</strong> the value of an element in <code>nums</code>.</li>\n\t<li>Calculate the <strong>sum</strong> of the elements of <code>nums</code> between indices <code>left</code> and <code>right</code> <strong>inclusive</strong> where <code>left &lt;= right</code>.</li>\n</ol>\n\n<p>Implement the <code>NumArray</code> class:</p>\n\n<ul>\n\t<li><code>NumArray(int[] nums)</code> Initializes the object with the integer array <code>nums</code>.</li>\n\t<li><code>void update(int index, int val)</code> <strong>Updates</strong> the value of <code>nums[index]</code> to be <code>val</code>.</li>\n\t<li><code>int sumRange(int left, int right)</code> Returns the <strong>sum</strong> of the elements of <code>nums</code> between indices <code>left</code> and <code>right</code> <strong>inclusive</strong> (i.e. <code>nums[left] + nums[left + 1] + ... + nums[right]</code>).</li>\n</ul>\n\n<p>&nbsp;</p>\n<p><strong>Example 1:</strong></p>\n\n<pre>\n<strong>Input</strong>\n[&quot;NumArray&quot;, &quot;sumRange&quot;, &quot;update&quot;, &quot;sumRange&quot;]\n[[[1, 3, 5]], [0, 2], [1, 2], [0, 2]]\n<strong>Output</strong>\n[null, 9, null, 8]\n\n<strong>Explanation</strong>\nNumArray numArray = new NumArray([1, 3, 5]);\nnumArray.sumRange(0, 2); // return 1 + 3 + 5 = 9\nnumArray.update(1, 2); // nums = [1, 2, 5]\nnumArray.sumRange(0, 2); // return 1 + 2 + 5 = 8\n</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= nums.length &lt;= 3 * 10<sup>4</sup></code></li>\n\t<li><code>-100 &lt;= nums[i] &lt;= 100</code></li>\n\t<li><code>0 &lt;= index &lt; nums.length</code></li>\n\t<li><code>-100 &lt;= val &lt;= 100</code></li>\n\t<li><code>0 &lt;= left &lt;= right &lt; nums.length</code></li>\n\t<li>At most <code>3 * 10<sup>4</sup></code> calls will be made to <code>update</code> and <code>sumRange</code>.</li>\n</ul>\n",
"translatedTitle": "区域和检索 - 数组可修改",
"translatedContent": "<p>给你一个数组 <code>nums</code> ,请你完成两类查询。</p>\n\n<ol>\n\t<li>其中一类查询要求 <strong>更新</strong> 数组&nbsp;<code>nums</code>&nbsp;下标对应的值</li>\n\t<li>另一类查询要求返回数组&nbsp;<code>nums</code>&nbsp;中索引&nbsp;<code>left</code>&nbsp;和索引&nbsp;<code>right</code>&nbsp;之间(&nbsp;<strong>包含&nbsp;</strong>的nums元素的 <strong>和</strong>&nbsp;,其中&nbsp;<code>left &lt;= right</code></li>\n</ol>\n\n<p>实现 <code>NumArray</code> 类:</p>\n\n<ul>\n\t<li><code>NumArray(int[] nums)</code> 用整数数组 <code>nums</code> 初始化对象</li>\n\t<li><code>void update(int index, int val)</code> 将 <code>nums[index]</code> 的值 <strong>更新</strong> 为 <code>val</code></li>\n\t<li><code>int sumRange(int left, int right)</code> 返回数组&nbsp;<code>nums</code>&nbsp;中索引&nbsp;<code>left</code>&nbsp;和索引&nbsp;<code>right</code>&nbsp;之间(&nbsp;<strong>包含&nbsp;</strong>的nums元素的 <strong>和</strong>&nbsp;(即,<code>nums[left] + nums[left + 1], ..., nums[right]</code></li>\n</ul>\n\n<p>&nbsp;</p>\n\n<p><strong>示例 1</strong></p>\n\n<pre>\n<strong>输入</strong>\n[\"NumArray\", \"sumRange\", \"update\", \"sumRange\"]\n[[[1, 3, 5]], [0, 2], [1, 2], [0, 2]]\n<strong>输出</strong>\n[null, 9, null, 8]\n\n<strong>解释</strong>\nNumArray numArray = new NumArray([1, 3, 5]);\nnumArray.sumRange(0, 2); // 返回 1 + 3 + 5 = 9\nnumArray.update(1, 2); // nums = [1,2,5]\nnumArray.sumRange(0, 2); // 返回 1 + 2 + 5 = 8\n</pre>\n\n<p>&nbsp;</p>\n\n<p><strong>提示:</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= nums.length &lt;= 3 *&nbsp;10<sup>4</sup></code></li>\n\t<li><code>-100 &lt;= nums[i] &lt;= 100</code></li>\n\t<li><code>0 &lt;= index &lt; nums.length</code></li>\n\t<li><code>-100 &lt;= val &lt;= 100</code></li>\n\t<li><code>0 &lt;= left &lt;= right &lt; nums.length</code></li>\n\t<li>调用 <code>pdate</code> 和 <code>sumRange</code> 方法次数不大于&nbsp;<code>3 * 10<sup>4</sup></code>&nbsp;</li>\n</ul>\n",
"translatedContent": "<p>给你一个数组 <code>nums</code> ,请你完成两类查询。</p>\n\n<ol>\n\t<li>其中一类查询要求 <strong>更新</strong> 数组&nbsp;<code>nums</code>&nbsp;下标对应的值</li>\n\t<li>另一类查询要求返回数组&nbsp;<code>nums</code>&nbsp;中索引&nbsp;<code>left</code>&nbsp;和索引&nbsp;<code>right</code>&nbsp;之间(&nbsp;<strong>包含&nbsp;</strong>的nums元素的 <strong>和</strong>&nbsp;,其中&nbsp;<code>left &lt;= right</code></li>\n</ol>\n\n<p>实现 <code>NumArray</code> 类:</p>\n\n<ul>\n\t<li><code>NumArray(int[] nums)</code> 用整数数组 <code>nums</code> 初始化对象</li>\n\t<li><code>void update(int index, int val)</code> 将 <code>nums[index]</code> 的值 <strong>更新</strong> 为 <code>val</code></li>\n\t<li><code>int sumRange(int left, int right)</code> 返回数组&nbsp;<code>nums</code>&nbsp;中索引&nbsp;<code>left</code>&nbsp;和索引&nbsp;<code>right</code>&nbsp;之间(&nbsp;<strong>包含&nbsp;</strong>的nums元素的 <strong>和</strong>&nbsp;(即,<code>nums[left] + nums[left + 1], ..., nums[right]</code></li>\n</ul>\n\n<p>&nbsp;</p>\n\n<p><strong>示例 1</strong></p>\n\n<pre>\n<strong>输入</strong>\n[\"NumArray\", \"sumRange\", \"update\", \"sumRange\"]\n[[[1, 3, 5]], [0, 2], [1, 2], [0, 2]]\n<strong>输出</strong>\n[null, 9, null, 8]\n\n<strong>解释</strong>\nNumArray numArray = new NumArray([1, 3, 5]);\nnumArray.sumRange(0, 2); // 返回 1 + 3 + 5 = 9\nnumArray.update(1, 2); // nums = [1,2,5]\nnumArray.sumRange(0, 2); // 返回 1 + 2 + 5 = 8\n</pre>\n\n<p>&nbsp;</p>\n\n<p><strong>提示:</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= nums.length &lt;= 3 *&nbsp;10<sup>4</sup></code></li>\n\t<li><code>-100 &lt;= nums[i] &lt;= 100</code></li>\n\t<li><code>0 &lt;= index &lt; nums.length</code></li>\n\t<li><code>-100 &lt;= val &lt;= 100</code></li>\n\t<li><code>0 &lt;= left &lt;= right &lt; nums.length</code></li>\n\t<li>调用 <code>update</code> 和 <code>sumRange</code> 方法次数不大于&nbsp;<code>3 * 10<sup>4</sup></code>&nbsp;</li>\n</ul>\n",
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