range-sum-query-mutable

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Algorithms

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

<ol>
	<li>Update the value of an element in <code>nums</code>.</li>
	<li>Calculate the sum of the elements of <code>nums</code> between indices <code>left</code> and <code>right</code> inclusive where <code>left &lt;= right</code>.</li>
</ol>

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

<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> Updates the value of <code>nums[index]</code> to be <code>val</code>.</li>
	<li><code>int sumRange(int left, int right)</code> Returns the sum of the elements of <code>nums</code> between indices <code>left</code> and <code>right</code> inclusive (i.e. <code>nums[left] + nums[left + 1] + ... + nums[right]</code>).</li>
</ul>

&nbsp;
Example 1:

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

Explanation
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>

&nbsp;
Constraints:

<ul>
	<li><code>1 &lt;= nums.length &lt;= 3 * 104</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 * 104</code> calls will be made to <code>update</code> and <code>sumRange</code>.</li>
</ul>

Range Sum Query - Mutable

Range Sum Query - Immutable

range-sum-query-immutable

区域和检索 - 数组不可变

Range Sum Query 2D - Mutable

range-sum-query-2d-mutable

二维区域和检索 - 矩阵可修改

Shifting Letters II

shifting-letters-ii

字母移位 II

Design

Binary Indexed Tree

Segment Tree

Array

给你一个数组 <code>nums</code> ，请你完成两类查询。

<ol>
	<li>其中一类查询要求 更新 数组&nbsp;<code>nums</code>&nbsp;下标对应的值</li>
	<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;包含&nbsp;）的nums元素的 和&nbsp;，其中&nbsp;<code>left &lt;= right</code></li>
</ol>

实现 <code>NumArray</code> 类：

<ul>
	<li><code>NumArray(int[] nums)</code> 用整数数组 <code>nums</code> 初始化对象</li>
	<li><code>void update(int index, int val)</code> 将 <code>nums[index]</code> 的值 更新 为 <code>val</code></li>
	<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;包含&nbsp;）的nums元素的 和&nbsp;（即，<code>nums[left] + nums[left + 1], ..., nums[right]</code>）</li>
</ul>

&nbsp;

示例 1：

<pre>
输入：
["NumArray", "sumRange", "update", "sumRange"]
[[[1, 3, 5]], [0, 2], [1, 2], [0, 2]]
输出：
[null, 9, null, 8]

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

&nbsp;

提示：

<ul>
	<li><code>1 &lt;= nums.length &lt;= 3 *&nbsp;104</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>调用 <code>update</code> 和 <code>sumRange</code> 方法次数不大于&nbsp;<code>3 * 104</code>&nbsp;</li>
</ul>