From ac30179a33a81b8bad7ea2b351cb4f5354ccfd37 Mon Sep 17 00:00:00 2001
From: huangge1199 <huangge1199@hotmail.com>
Date: Mon, 7 Jun 2021 10:43:55 +0800
Subject: [PATCH] =?UTF-8?q?307:=E5=8C=BA=E5=9F=9F=E5=92=8C=E6=A3=80?=
 =?UTF-8?q?=E7=B4=A2=20-=20=E6=95=B0=E7=BB=84=E5=8F=AF=E4=BF=AE=E6=94=B9?=
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---
 .../editor/cn/RangeSumQueryMutable.java       | 90 +++++++++++++++++++
 .../editor/cn/RangeSumQueryMutable.md         | 45 ++++++++++
 2 files changed, 135 insertions(+)
 create mode 100644 src/main/java/leetcode/editor/cn/RangeSumQueryMutable.java
 create mode 100644 src/main/java/leetcode/editor/cn/RangeSumQueryMutable.md

diff --git a/src/main/java/leetcode/editor/cn/RangeSumQueryMutable.java b/src/main/java/leetcode/editor/cn/RangeSumQueryMutable.java
new file mode 100644
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+++ b/src/main/java/leetcode/editor/cn/RangeSumQueryMutable.java
@@ -0,0 +1,90 @@
+//给你一个数组 nums ，请你完成两类查询，其中一类查询要求更新数组下标对应的值，另一类查询要求返回数组中某个范围内元素的总和。 
+//
+// 实现 NumArray 类： 
+//
+// 
+// 
+// 
+// NumArray(int[] nums) 用整数数组 nums 初始化对象 
+// void update(int index, int val) 将 nums[index] 的值更新为 val 
+// int sumRange(int left, int right) 返回子数组 nums[left, right] 的总和（即，nums[left] + 
+//nums[left + 1], ..., nums[right]） 
+// 
+//
+// 
+//
+// 示例： 
+//
+// 
+//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 9 ，sum([1,3,5]) = 9
+//numArray.update(1, 2);   // nums = [1,2,5]
+//numArray.sumRange(0, 2); // 返回 8 ，sum([1,2,5]) = 8
+// 
+//
+// 
+//
+// 提示： 
+//
+// 
+// 1 <= nums.length <= 3 * 104 
+// -100 <= nums[i] <= 100 
+// 0 <= index < nums.length 
+// -100 <= val <= 100 
+// 0 <= left <= right < nums.length 
+// 最多调用 3 * 104 次 update 和 sumRange 方法 
+// 
+// 
+// 
+// Related Topics 树状数组 线段树 
+// 👍 273 👎 0
+
+package leetcode.editor.cn;
+//307:区域和检索 - 数组可修改
+public class RangeSumQueryMutable{
+    public static void main(String[] args) {
+        //测试代码
+//        Solution solution = new RangeSumQueryMutable().new Solution();
+    }
+    //力扣代码
+    //leetcode submit region begin(Prohibit modification and deletion)
+class NumArray {
+
+        private int[] nums;
+
+        public NumArray(int[] nums) {
+            this.nums = new int[nums.length];
+            for (int i = 0; i < nums.length; i++) {
+                update(i, nums[i]);
+            }
+        }
+
+        public void update(int i, int val) {
+            this.nums[i] = val;
+        }
+
+        public int sumRange(int i, int j) {
+            int sum = 0;
+            for (; i <= j; i++) {
+                sum += this.nums[i];
+            }
+            return sum;
+        }
+}
+
+/**
+ * Your NumArray object will be instantiated and called as such:
+ * NumArray obj = new NumArray(nums);
+ * obj.update(index,val);
+ * int param_2 = obj.sumRange(left,right);
+ */
+//leetcode submit region end(Prohibit modification and deletion)
+
+}
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diff --git a/src/main/java/leetcode/editor/cn/RangeSumQueryMutable.md b/src/main/java/leetcode/editor/cn/RangeSumQueryMutable.md
new file mode 100644
index 0000000..c5f9cbb
--- /dev/null
+++ b/src/main/java/leetcode/editor/cn/RangeSumQueryMutable.md
@@ -0,0 +1,45 @@
+<p>给你一个数组 <code>nums</code> ，请你完成两类查询，其中一类查询要求更新数组下标对应的值，另一类查询要求返回数组中某个范围内元素的总和。</p>
+
+<p>实现 <code>NumArray</code> 类：</p>
+
+<div class="original__bRMd">
+<div>
+<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> 返回子数组 <code>nums[left, right]</code> 的总和（即，<code>nums[left] + nums[left + 1], ..., nums[right]</code>）</li>
+</ul>
+
+<p> </p>
+
+<p><strong>示例：</strong></p>
+
+<pre>
+<strong>输入</strong>：
+["NumArray", "sumRange", "update", "sumRange"]
+[[[1, 3, 5]], [0, 2], [1, 2], [0, 2]]
+<strong>输出</strong>：
+[null, 9, null, 8]
+
+<strong>解释</strong>：
+NumArray numArray = new NumArray([1, 3, 5]);
+numArray.sumRange(0, 2); // 返回 9 ，sum([1,3,5]) = 9
+numArray.update(1, 2);   // nums = [1,2,5]
+numArray.sumRange(0, 2); // 返回 8 ，sum([1,2,5]) = 8
+</pre>
+
+<p> </p>
+
+<p><strong>提示：</strong></p>
+
+<ul>
+	<li><code>1 <= nums.length <= 3 * 10<sup>4</sup></code></li>
+	<li><code>-100 <= nums[i] <= 100</code></li>
+	<li><code>0 <= index < nums.length</code></li>
+	<li><code>-100 <= val <= 100</code></li>
+	<li><code>0 <= left <= right < nums.length</code></li>
+	<li>最多调用 <code>3 * 10<sup>4</sup></code> 次 <code>update</code> 和 <code>sumRange</code> 方法</li>
+</ul>
+</div>
+</div>
+<div><div>Related Topics</div><div><li>树状数组</li><li>线段树</li></div></div>\n<div><li>👍 273</li><li>👎 0</li></div>
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