mirror of
https://gitee.com/coder-xiaomo/leetcode-problemset
synced 2025-01-10 18:48:13 +08:00
45 lines
2.0 KiB
HTML
45 lines
2.0 KiB
HTML
<p>给你一个字符串数组 <code>queries</code>,和一个表示模式的字符串 <code>pattern</code>,请你返回一个布尔数组 <code>answer</code> 。只有在待查项 <code>queries[i]</code> 与模式串 <code>pattern</code> 匹配时, <code>answer[i]</code> 才为 <code>true</code>,否则为 <code>false</code>。</p>
|
||
|
||
<p>如果可以将<strong>小写字母</strong>插入模式串 <code>pattern</code> 得到待查询项 <code>query</code>,那么待查询项与给定模式串匹配。可以在任何位置插入每个字符,也可以不插入字符。</p>
|
||
|
||
<p> </p>
|
||
|
||
<p><strong>示例 1:</strong></p>
|
||
|
||
<pre>
|
||
<strong>输入:</strong>queries = ["FooBar","FooBarTest","FootBall","FrameBuffer","ForceFeedBack"], pattern = "FB"
|
||
<strong>输出:</strong>[true,false,true,true,false]
|
||
<strong>示例:</strong>
|
||
"FooBar" 可以这样生成:"F" + "oo" + "B" + "ar"。
|
||
"FootBall" 可以这样生成:"F" + "oot" + "B" + "all".
|
||
"FrameBuffer" 可以这样生成:"F" + "rame" + "B" + "uffer".</pre>
|
||
|
||
<p><strong>示例 2:</strong></p>
|
||
|
||
<pre>
|
||
<strong>输入:</strong>queries = ["FooBar","FooBarTest","FootBall","FrameBuffer","ForceFeedBack"], pattern = "FoBa"
|
||
<strong>输出:</strong>[true,false,true,false,false]
|
||
<strong>解释:</strong>
|
||
"FooBar" 可以这样生成:"Fo" + "o" + "Ba" + "r".
|
||
"FootBall" 可以这样生成:"Fo" + "ot" + "Ba" + "ll".
|
||
</pre>
|
||
|
||
<p><strong>示例 3:</strong></p>
|
||
|
||
<pre>
|
||
<strong>输入:</strong>queries = ["FooBar","FooBarTest","FootBall","FrameBuffer","ForceFeedBack"], pattern = "FoBaT"
|
||
<strong>输出:</strong>[false,true,false,false,false]
|
||
<strong>解释: </strong>
|
||
"FooBarTest" 可以这样生成:"Fo" + "o" + "Ba" + "r" + "T" + "est".
|
||
</pre>
|
||
|
||
<p> </p>
|
||
|
||
<p><strong>提示:</strong></p>
|
||
|
||
<ul>
|
||
<li><code>1 <= pattern.length, queries.length <= 100</code></li>
|
||
<li><code>1 <= queries[i].length <= 100</code></li>
|
||
<li><code>queries[i]</code> 和 <code>pattern</code> 由英文字母组成</li>
|
||
</ul>
|