leetocde前缀和

package editor.cn;

import java.util.Arrays;

/**
<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>

<ol>
	<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>
</ol>

<p>实现 <code>NumArray</code> 类：</p>

<ul>
	<li><code>NumArray(int[] nums)</code> 使用数组 <code>nums</code> 初始化对象</li>
	<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>
</ul>

<p>&nbsp;</p>

<p><strong>示例 1：</strong></p>

<pre>
<strong>输入：</strong>
["NumArray", "sumRange", "sumRange", "sumRange"]
[[[-2, 0, 3, -5, 2, -1]], [0, 2], [2, 5], [0, 5]]
<strong>输出：
</strong>[null, 1, -1, -3]

<strong>解释：</strong>
NumArray numArray = new NumArray([-2, 0, 3, -5, 2, -1]);
numArray.sumRange(0, 2); // return 1 ((-2) + 0 + 3)
numArray.sumRange(2, 5); // return -1 (3 + (-5) + 2 + (-1)) 
numArray.sumRange(0, 5); // return -3 ((-2) + 0 + 3 + (-5) + 2 + (-1))
</pre>

<p>&nbsp;</p>

<p><strong>提示：</strong></p>

<ul>
	<li><code>1 &lt;= nums.length &lt;= 10<sup>4</sup></code></li>
	<li><code>-10<sup>5</sup>&nbsp;&lt;= nums[i] &lt;=&nbsp;10<sup>5</sup></code></li>
	<li><code>0 &lt;= i &lt;= j &lt; nums.length</code></li>
	<li>最多调用 <code>10<sup>4</sup></code> 次 <code>sumRange</code><strong> </strong>方法</li>
</ul>
<div><div>Related Topics</div><div><li>设计</li><li>数组</li><li>前缀和</li></div></div><br><div><li> 422</li><li> 0</li></div>
*/

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class NumArray {

	// 考点：前缀和数组
	private int[] preSum;

	/* 输入一个数组，构造前缀和 */
	public NumArray(int[] nums) {
		// preSum[0] = 0，便于计算累加和
		preSum = new int[nums.length + 1];
		// 计算 nums 的累加和
		for (int i = 1; i < preSum.length; i++) {
			preSum[i] = preSum[i - 1] + nums[i - 1];
		}
	}

	/* 查询闭区间 [left, right] 的累加和 */
	//此处preSum_left代表 0-（left-1）的和
	//preSum-right 代表 0- (right-1)的和
	//查询闭区间 [left, right] 的累加和 等于 presum(right）-presum(left)+nums(right) = preSum[right + 1] - preSum[left];
	public int sumRange(int left, int right) {
		return preSum[right + 1] - preSum[left];
	}
}

/**
 * Your NumArray object will be instantiated and called as such:
 * NumArray obj = new NumArray(nums);
 * int param_1 = obj.sumRange(left,right);
 */
//leetcode submit region end(Prohibit modification and deletion)