mirror of
https://gitee.com/coder-xiaomo/leetcode-problemset
synced 2025-01-11 02:58:13 +08:00
66 lines
2.7 KiB
HTML
66 lines
2.7 KiB
HTML
|
<p>如果可以使用以下操作从一个字符串得到另一个字符串，则认为两个字符串 <strong>接近</strong> ：</p>
|
|||
|
|
|
|||
|
|
<ul>
|
|||
|
|
<li>操作 1：交换任意两个 <strong>现有</strong> 字符。
|
|||
|
|
|
|||
|
|
<ul>
|
|||
|
|
<li>例如，<code>a<strong>b</strong>cd<strong>e</strong> -> a<strong>e</strong>cd<strong>b</strong></code></li>
|
|||
|
|
</ul>
|
|||
|
|
</li>
|
|||
|
|
<li>操作 2：将一个 <strong>现有</strong> 字符的每次出现转换为另一个 <strong>现有</strong> 字符，并对另一个字符执行相同的操作。
|
|||
|
|
<ul>
|
|||
|
|
<li>例如，<code><strong>aa</strong>c<strong>abb</strong> -> <strong>bb</strong>c<strong>baa</strong></code>（所有 <code>a</code> 转化为 <code>b</code> ，而所有的 <code>b</code> 转换为 <code>a</code> ）</li>
|
|||
|
|
</ul>
|
|||
|
|
</li>
|
|||
|
|
</ul>
|
|||
|
|
|
|||
|
|
<p>你可以根据需要对任意一个字符串多次使用这两种操作。</p>
|
|||
|
|
|
|||
|
|
<p>给你两个字符串，<code>word1</code> 和 <code>word2</code> 。如果<em> </em><code>word1</code><em> </em>和<em> </em><code>word2</code><em> </em><strong>接近 </strong>，就返回 <code>true</code> ；否则，返回<em> </em><code>false</code><em> </em>。</p>
|
|||
|
|
|
|||
|
|
<p> </p>
|
|||
|
|
|
|||
|
|
<p><strong>示例 1：</strong></p>
|
|||
|
|
|
|||
|
|
<pre>
|
|||
|
|
<strong>输入：</strong>word1 = "abc", word2 = "bca"
|
|||
|
|
<strong>输出：</strong>true
|
|||
|
|
<strong>解释：</strong>2 次操作从 word1 获得 word2 。
|
|||
|
|
执行操作 1："a<strong>bc</strong>" -> "a<strong>cb</strong>"
|
|||
|
|
执行操作 1："<strong>a</strong>c<strong>b</strong>" -> "<strong>b</strong>c<strong>a</strong>"
|
|||
|
|
</pre>
|
|||
|
|
|
|||
|
|
<p><strong>示例 2：</strong></p>
|
|||
|
|
|
|||
|
|
<pre>
|
|||
|
|
<strong>输入：</strong>word1 = "a", word2 = "aa"
|
|||
|
|
<strong>输出：</strong>false
|
|||
|
|
<strong>解释：</strong>不管执行多少次操作，都无法从 word1 得到 word2 ，反之亦然。</pre>
|
|||
|
|
|
|||
|
|
<p><strong>示例 3：</strong></p>
|
|||
|
|
|
|||
|
|
<pre>
|
|||
|
|
<strong>输入：</strong>word1 = "cabbba", word2 = "abbccc"
|
|||
|
|
<strong>输出：</strong>true
|
|||
|
|
<strong>解释：</strong>3 次操作从 word1 获得 word2 。
|
|||
|
|
执行操作 1："ca<strong>b</strong>bb<strong>a</strong>" -> "ca<strong>a</strong>bb<strong>b</strong>"
|
|||
|
|
执行操作 2：<code>"</code><strong>c</strong>aa<strong>bbb</strong>" -> "<strong>b</strong>aa<strong>ccc</strong>"
|
|||
|
|
执行操作 2："<strong>baa</strong>ccc" -> "<strong>abb</strong>ccc"
|
|||
|
|
</pre>
|
|||
|
|
|
|||
|
|
<p><strong>示例 4：</strong></p>
|
|||
|
|
|
|||
|
|
<pre>
|
|||
|
|
<strong>输入：</strong>word1 = "cabbba", word2 = "aabbss"
|
|||
|
|
<strong>输出：</strong>false
|
|||
|
|
<strong>解释：</strong>不管执行多少次操作，都无法从 word1 得到 word2 ，反之亦然。</pre>
|
|||
|
|
|
|||
|
|
<p> </p>
|
|||
|
|
|
|||
|
|
<p><strong>提示：</strong></p>
|
|||
|
|
|
|||
|
|
<ul>
|
|||
|
|
<li><code>1 <= word1.length, word2.length <= 10<sup>5</sup></code></li>
|
|||
|
|
<li><code>word1</code> 和 <code>word2</code> 仅包含小写英文字母</li>
|
|||
|
|
</ul>
|