leetcode-problemset/leetcode-cn/problem (English)/区域和检索 - 数组可修改(English) [range-sum-query-mutable].html

<p>Given an integer array <code>nums</code>, handle multiple queries of the following types:</p>

<ol>
	<li><strong>Update</strong> the value of an element in <code>nums</code>.</li>
	<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>
</ol>

<p>Implement the <code>NumArray</code> class:</p>

<ul>
	<li><code>NumArray(int[] nums)</code> Initializes the object with the integer array <code>nums</code>.</li>
	<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>
	<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>
</ul>

<p>&nbsp;</p>
<p><strong>Example 1:</strong></p>

<pre>
<strong>Input</strong>
[&quot;NumArray&quot;, &quot;sumRange&quot;, &quot;update&quot;, &quot;sumRange&quot;]
[[[1, 3, 5]], [0, 2], [1, 2], [0, 2]]
<strong>Output</strong>
[null, 9, null, 8]

<strong>Explanation</strong>
NumArray numArray = new NumArray([1, 3, 5]);
numArray.sumRange(0, 2); // return 1 + 3 + 5 = 9
numArray.update(1, 2);   // nums = [1, 2, 5]
numArray.sumRange(0, 2); // return 1 + 2 + 5 = 8
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums.length &lt;= 3 * 10<sup>4</sup></code></li>
	<li><code>-100 &lt;= nums[i] &lt;= 100</code></li>
	<li><code>0 &lt;= index &lt; nums.length</code></li>
	<li><code>-100 &lt;= val &lt;= 100</code></li>
	<li><code>0 &lt;= left &lt;= right &lt; nums.length</code></li>
	<li>At most <code>3 * 10<sup>4</sup></code> calls will be made to <code>update</code> and <code>sumRange</code>.</li>
</ul>