mirror of
https://gitee.com/coder-xiaomo/leetcode-problemset
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168 lines
25 KiB
JSON
168 lines
25 KiB
JSON
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"title": "String to Integer (atoi)",
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"content": "<p>Implement the <code>myAtoi(string s)</code> function, which converts a string to a 32-bit signed integer (similar to C/C++'s <code>atoi</code> function).</p>\n\n<p>The algorithm for <code>myAtoi(string s)</code> is as follows:</p>\n\n<ol>\n\t<li>Read in and ignore any leading whitespace.</li>\n\t<li>Check if the next character (if not already at the end of the string) is <code>'-'</code> or <code>'+'</code>. Read this character in if it is either. This determines if the final result is negative or positive respectively. Assume the result is positive if neither is present.</li>\n\t<li>Read in next the characters until the next non-digit character or the end of the input is reached. The rest of the string is ignored.</li>\n\t<li>Convert these digits into an integer (i.e. <code>"123" -> 123</code>, <code>"0032" -> 32</code>). If no digits were read, then the integer is <code>0</code>. Change the sign as necessary (from step 2).</li>\n\t<li>If the integer is out of the 32-bit signed integer range <code>[-2<sup>31</sup>, 2<sup>31</sup> - 1]</code>, then clamp the integer so that it remains in the range. Specifically, integers less than <code>-2<sup>31</sup></code> should be clamped to <code>-2<sup>31</sup></code>, and integers greater than <code>2<sup>31</sup> - 1</code> should be clamped to <code>2<sup>31</sup> - 1</code>.</li>\n\t<li>Return the integer as the final result.</li>\n</ol>\n\n<p><strong>Note:</strong></p>\n\n<ul>\n\t<li>Only the space character <code>' '</code> is considered a whitespace character.</li>\n\t<li><strong>Do not ignore</strong> any characters other than the leading whitespace or the rest of the string after the digits.</li>\n</ul>\n\n<p> </p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> s = "42"\n<strong>Output:</strong> 42\n<strong>Explanation:</strong> The underlined characters are what is read in, the caret is the current reader position.\nStep 1: "42" (no characters read because there is no leading whitespace)\n ^\nStep 2: "42" (no characters read because there is neither a '-' nor '+')\n ^\nStep 3: "<u>42</u>" ("42" is read in)\n ^\nThe parsed integer is 42.\nSince 42 is in the range [-2<sup>31</sup>, 2<sup>31</sup> - 1], the final result is 42.\n</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> s = " -42"\n<strong>Output:</strong> -42\n<strong>Explanation:</strong>\nStep 1: "<u> </u>-42" (leading whitespace is read and ignored)\n ^\nStep 2: " <u>-</u>42" ('-' is read, so the result should be negative)\n ^\nStep 3: " -<u>42</u>" ("42" is read in)\n ^\nThe parsed integer is -42.\nSince -42 is in the range [-2<sup>31</sup>, 2<sup>31</sup> - 1], the final result is -42.\n</pre>\n\n<p><strong class=\"example\">Example 3:</strong></p>\n\n<pre>\n<strong>Input:</strong> s = "4193 with words"\n<strong>Output:</strong> 4193\n<strong>Explanation:</strong>\nStep 1: "4193 with words" (no characters read because there is no leading whitespace)\n ^\nStep 2: "4193 with words" (no characters read because there is neither a '-' nor '+')\n ^\nStep 3: "<u>4193</u> with words" ("4193" is read in; reading stops because the next character is a non-digit)\n ^\nThe parsed integer is 4193.\nSince 4193 is in the range [-2<sup>31</sup>, 2<sup>31</sup> - 1], the final result is 4193.\n</pre>\n\n<p> </p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>0 <= s.length <= 200</code></li>\n\t<li><code>s</code> consists of English letters (lower-case and upper-case), digits (<code>0-9</code>), <code>' '</code>, <code>'+'</code>, <code>'-'</code>, and <code>'.'</code>.</li>\n</ul>\n",
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"translatedTitle": "字符串转换整数 (atoi)",
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"translatedContent": "<p>请你来实现一个 <code>myAtoi(string s)</code> 函数,使其能将字符串转换成一个 32 位有符号整数(类似 C/C++ 中的 <code>atoi</code> 函数)。</p>\n\n<p>函数 <code>myAtoi(string s)</code> 的算法如下:</p>\n\n<ol>\n\t<li>读入字符串并丢弃无用的前导空格</li>\n\t<li>检查下一个字符(假设还未到字符末尾)为正还是负号,读取该字符(如果有)。 确定最终结果是负数还是正数。 如果两者都不存在,则假定结果为正。</li>\n\t<li>读入下一个字符,直到到达下一个非数字字符或到达输入的结尾。字符串的其余部分将被忽略。</li>\n\t<li>将前面步骤读入的这些数字转换为整数(即,\"123\" -> 123, \"0032\" -> 32)。如果没有读入数字,则整数为 <code>0</code> 。必要时更改符号(从步骤 2 开始)。</li>\n\t<li>如果整数数超过 32 位有符号整数范围 <code>[−2<sup>31</sup>, 2<sup>31 </sup>− 1]</code> ,需要截断这个整数,使其保持在这个范围内。具体来说,小于 <code>−2<sup>31</sup></code> 的整数应该被固定为 <code>−2<sup>31</sup></code> ,大于 <code>2<sup>31 </sup>− 1</code> 的整数应该被固定为 <code>2<sup>31 </sup>− 1</code> 。</li>\n\t<li>返回整数作为最终结果。</li>\n</ol>\n\n<p><strong>注意:</strong></p>\n\n<ul>\n\t<li>本题中的空白字符只包括空格字符 <code>' '</code> 。</li>\n\t<li>除前导空格或数字后的其余字符串外,<strong>请勿忽略</strong> 任何其他字符。</li>\n</ul>\n\n<p> </p>\n\n<p><strong>示例 1:</strong></p>\n\n<pre>\n<strong>输入:</strong>s = \"42\"\n<strong>输出:</strong>42\n<strong>解释:</strong>加粗的字符串为已经读入的字符,插入符号是当前读取的字符。\n第 1 步:\"42\"(当前没有读入字符,因为没有前导空格)\n ^\n第 2 步:\"42\"(当前没有读入字符,因为这里不存在 '-' 或者 '+')\n ^\n第 3 步:\"<u>42</u>\"(读入 \"42\")\n ^\n解析得到整数 42 。\n由于 \"42\" 在范围 [-2<sup>31</sup>, 2<sup>31</sup> - 1] 内,最终结果为 42 。</pre>\n\n<p><strong>示例 2:</strong></p>\n\n<pre>\n<strong>输入:</strong>s = \" -42\"\n<strong>输出:</strong>-42\n<strong>解释:</strong>\n第 1 步:\"<u><strong> </strong></u>-42\"(读入前导空格,但忽视掉)\n ^\n第 2 步:\" <u><strong>-</strong></u>42\"(读入 '-' 字符,所以结果应该是负数)\n ^\n第 3 步:\" <u><strong>-42</strong></u>\"(读入 \"42\")\n ^\n解析得到整数 -42 。\n由于 \"-42\" 在范围 [-2<sup>31</sup>, 2<sup>31</sup> - 1] 内,最终结果为 -42 。\n</pre>\n\n<p><strong>示例 3:</strong></p>\n\n<pre>\n<strong>输入:</strong>s = \"4193 with words\"\n<strong>输出:</strong>4193\n<strong>解释:</strong>\n第 1 步:\"4193 with words\"(当前没有读入字符,因为没有前导空格)\n ^\n第 2 步:\"4193 with words\"(当前没有读入字符,因为这里不存在 '-' 或者 '+')\n ^\n第 3 步:\"<u>4193</u> with words\"(读入 \"4193\";由于下一个字符不是一个数字,所以读入停止)\n ^\n解析得到整数 4193 。\n由于 \"4193\" 在范围 [-2<sup>31</sup>, 2<sup>31</sup> - 1] 内,最终结果为 4193 。\n</pre>\n\n<p> </p>\n\n<p><strong>提示:</strong></p>\n\n<ul>\n\t<li><code>0 <= s.length <= 200</code></li>\n\t<li><code>s</code> 由英文字母(大写和小写)、数字(<code>0-9</code>)、<code>' '</code>、<code>'+'</code>、<code>'-'</code> 和 <code>'.'</code> 组成</li>\n</ul>\n",
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