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leetcode-problemset/leetcode-cn/originData/count-odd-numbers-in-an-interval-range.json
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{
"data": {
"question": {
"questionId": "1630",
"questionFrontendId": "1523",
"categoryTitle": "Algorithms",
"boundTopicId": 344041,
"title": "Count Odd Numbers in an Interval Range",
"titleSlug": "count-odd-numbers-in-an-interval-range",
"content": "<p>Given two non-negative integers <code>low</code> and <code><font face=\"monospace\">high</font></code>. Return the <em>count of odd numbers between </em><code>low</code><em> and </em><code><font face=\"monospace\">high</font></code><em>&nbsp;(inclusive)</em>.</p>\r\n\r\n<p>&nbsp;</p>\r\n<p><strong class=\"example\">Example 1:</strong></p>\r\n\r\n<pre>\r\n<strong>Input:</strong> low = 3, high = 7\r\n<strong>Output:</strong> 3\r\n<b>Explanation: </b>The odd numbers between 3 and 7 are [3,5,7].</pre>\r\n\r\n<p><strong class=\"example\">Example 2:</strong></p>\r\n\r\n<pre>\r\n<strong>Input:</strong> low = 8, high = 10\r\n<strong>Output:</strong> 1\r\n<b>Explanation: </b>The odd numbers between 8 and 10 are [9].</pre>\r\n\r\n<p>&nbsp;</p>\r\n<p><strong>Constraints:</strong></p>\r\n\r\n<ul>\r\n\t<li><code>0 &lt;= low &lt;= high&nbsp;&lt;= 10^9</code></li>\r\n</ul>",
"translatedTitle": "在区间范围内统计奇数数目",
"translatedContent": "<p>给你两个非负整数&nbsp;<code>low</code> 和&nbsp;<code>high</code>&nbsp;。请你返回<em>&nbsp;</em><code>low</code><em> </em>和<em>&nbsp;</em><code>high</code><em>&nbsp;</em>之间(包括二者)奇数的数目。</p>\n\n<p>&nbsp;</p>\n\n<p><strong>示例 1</strong></p>\n\n<pre><strong>输入:</strong>low = 3, high = 7\n<strong>输出:</strong>3\n<strong>解释:</strong>3 到 7 之间奇数数字为 [3,5,7] 。</pre>\n\n<p><strong>示例 2</strong></p>\n\n<pre><strong>输入:</strong>low = 8, high = 10\n<strong>输出:</strong>1\n<strong>解释:</strong>8 到 10 之间奇数数字为 [9] 。</pre>\n\n<p>&nbsp;</p>\n\n<p><strong>提示:</strong></p>\n\n<ul>\n\t<li><code>0 &lt;= low &lt;= high&nbsp;&lt;= 10^9</code></li>\n</ul>\n",
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"If the range (high - low + 1) is even, the number of even and odd numbers in this range will be the same.",
"If the range (high - low + 1) is odd, the solution will depend on the parity of high and low."
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