leetcode-problemset/leetcode-cn/originData/range-sum-query-immutable.json

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        "question": {
            "questionId": "303",
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            "categoryTitle": "Algorithms",
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            "title": "Range Sum Query - Immutable",
            "titleSlug": "range-sum-query-immutable",
            "content": "<p>Given an integer array <code>nums</code>, handle multiple queries of the following type:</p>\n\n<ol>\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>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 class=\"example\">Example 1:</strong></p>\n\n<pre>\n<strong>Input</strong>\n[&quot;NumArray&quot;, &quot;sumRange&quot;, &quot;sumRange&quot;, &quot;sumRange&quot;]\n[[[-2, 0, 3, -5, 2, -1]], [0, 2], [2, 5], [0, 5]]\n<strong>Output</strong>\n[null, 1, -1, -3]\n\n<strong>Explanation</strong>\nNumArray numArray = new NumArray([-2, 0, 3, -5, 2, -1]);\nnumArray.sumRange(0, 2); // return (-2) + 0 + 3 = 1\nnumArray.sumRange(2, 5); // return 3 + (-5) + 2 + (-1) = -1\nnumArray.sumRange(0, 5); // return (-2) + 0 + 3 + (-5) + 2 + (-1) = -3\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;= 10<sup>4</sup></code></li>\n\t<li><code>-10<sup>5</sup> &lt;= nums[i] &lt;= 10<sup>5</sup></code></li>\n\t<li><code>0 &lt;= left &lt;= right &lt; nums.length</code></li>\n\t<li>At most <code>10<sup>4</sup></code> calls will be made to <code>sumRange</code>.</li>\n</ul>\n",
            "translatedTitle": "区域和检索 - 数组不可变",
            "translatedContent": "<p>给定一个整数数组 &nbsp;<code>nums</code>，处理以下类型的多个查询:</p>\n\n<ol>\n\t<li>计算索引&nbsp;<code>left</code>&nbsp;和&nbsp;<code>right</code>&nbsp;（包含 <code>left</code> 和 <code>right</code>）之间的 <code>nums</code> 元素的 <strong>和</strong> ，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> 使用数组 <code>nums</code> 初始化对象</li>\n\t<li><code>int sumRange(int i, int j)</code> 返回数组 <code>nums</code>&nbsp;中索引&nbsp;<code>left</code>&nbsp;和&nbsp;<code>right</code>&nbsp;之间的元素的 <strong>总和</strong> ，包含&nbsp;<code>left</code>&nbsp;和&nbsp;<code>right</code>&nbsp;两点（也就是&nbsp;<code>nums[left] + nums[left + 1] + ... + nums[right]</code>&nbsp;)</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\", \"sumRange\", \"sumRange\"]\n[[[-2, 0, 3, -5, 2, -1]], [0, 2], [2, 5], [0, 5]]\n<strong>输出：\n</strong>[null, 1, -1, -3]\n\n<strong>解释：</strong>\nNumArray numArray = new NumArray([-2, 0, 3, -5, 2, -1]);\nnumArray.sumRange(0, 2); // return 1 ((-2) + 0 + 3)\nnumArray.sumRange(2, 5); // return -1 (3 + (-5) + 2 + (-1)) \nnumArray.sumRange(0, 5); // return -3 ((-2) + 0 + 3 + (-5) + 2 + (-1))\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;= 10<sup>4</sup></code></li>\n\t<li><code>-10<sup>5</sup>&nbsp;&lt;= nums[i] &lt;=&nbsp;10<sup>5</sup></code></li>\n\t<li><code>0 &lt;= i &lt;= j &lt; nums.length</code></li>\n\t<li>最多调用 <code>10<sup>4</sup></code> 次 <code>sumRange</code><strong> </strong>方法</li>\n</ul>\n",
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