307:区域和检索 - 数组可修改

This commit is contained in:
huangge1199 2021-06-07 10:43:55 +08:00
parent badb2abbf5
commit ac30179a33
2 changed files with 135 additions and 0 deletions

View File

@ -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
//
//
//
//
// NumArray(int[] nums) 用整数数组 nums 初始化对象
// void update(int index, int val) 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 update 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)
}

View File

@ -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><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><code>update</code><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>