编码和解码字符串

2026-08-02T00:00:00+00:00
题目：实现字符串数组到字符串的序列化和反序列化方法，使得输入的字符串数组经过一次编码和一次解码后，仍和原来保持一致。</p>
string</span> encode</span>(</span>vector</span><</span>string</span>></span>&</span> strs</span>) {}</span></span>
</span>
vector</span><</span>string</span>></span> decode</span>(</span>string</span> s</span>) {}</span>  </span></span></code></pre>题解</h1>
遍历字符串数组，计算每个元素的长度，增加到字符串开头，并增加一个特殊界定符，用于隔开长度和原始字符串：str.size() + '#' + str</code>。</p>
字符串中出现了 '#' 符号会不会有影响？</p>
答案是不会，关键在于 '#' 第一次出现的位置，这是可以确定的，然后将位置左边的内容解析为长度值，位置往右数长度值个字符的内容解析为原始字符串。</p>
看一个例子就比较清晰了：</p>
5 # h e # # o 5 # w o r l d</span></span>
  ^             ^</span></span>
  首个 # 位置   </span></span></code></pre>
找到首个 # 位置，解析长度为 5 ，往 # 位置后数 5 个字符，解析字符串为 "he##o"。也就是说，原始字符串中存在 # 符号也不会有任何影响，会被完整解析为字符串。</p>
依此类推直到最后一个字符串。这里需要使用的一个技巧是：解析完第一个字符串后，将这部分内容删除，使得解析后续字符串都如同首个一样。</p>
代码</h1>
string</span> encode</span>(</span>vector</span><</span>string</span>></span>&</span> strs</span>) {</span></span>
    string ret </span>=</span> ""</span>;</span></span>
    for</span> (</span>string str </span>:</span> strs</span>) {</span></span>
        ret </span>+=</span> to_string</span>(</span>str</span>.</span>length</span>())</span> +</span> "#"</span> +</span> str</span>;</span> </span></span>
    }</span></span>
    return</span> ret</span>;</span></span>
}</span></span>
</span>
vector</span><</span>string</span>></span> decode</span>(</span>string</span> s</span>) {</span></span>
    vector</span><</span>string</span>></span> ret </span>=</span> {};</span></span>
    while</span> (</span>!</span>s</span>.</span>empty</span>()) {</span></span>
        if</span> (</span>s</span>.</span>find</span>(</span>"#"</span>)</span> !=</span> string</span>::</span>npos</span>) {</span></span>
            int</span> pos </span>=</span> s</span>.</span>find</span>(</span>"#"</span>);</span></span>
            size_t</span> len </span>=</span> stoi</span>(</span>s</span>.</span>substr</span>(</span>0</span>,</span> pos</span>));</span></span>
            ret</span>.</span>push_back</span>(</span>s</span>.</span>substr</span>(</span>pos </span>+</span> 1</span>,</span> len</span>));</span></span>
            s</span>.</span>erase</span>(</span>0</span>,</span> s</span>.</span>substr</span>(</span>0</span>,</span> pos</span>).</span>length</span>()</span> +</span> 1</span> +</span> len</span>);</span></span>
        }</span></span>
    }</span></span>
    return</span> ret</span>;</span></span>
}</span></span></code></pre>历程</h1>
相较于其他算法题，这更像是一个程序员日常需要解决的工程问题。</p>
一上来没想那么多，直接使用逗号分隔符号连接字符串数组。很明显，这种方法不能通过测试用例包含逗号分隔符的情况，失败。</p>
查看提示，了解到可以通过记录字符串长度进行标识，如何记录字符串长度信息成为关键。但我马上又遇到新的问题；长度本身也占不定长度字符，不好解析。</p>
再次审题，发现每个字符串的长度不会超过 200，也就是说三位数字就足以存储所有可能的字符串（不满三位时补充0）。在解析字符串时，前三位固定为长度值，紧接着为原始字符串。很幸运，通过测试。</p>
反思，但如果字符串是任意长度，代码就不适应了。查看答案，发现只要在长度和字符串之间增加一个特定符号就可以解决。妙！</p>
链接</h1>

https://neetcode.io/problems/string-encode-and-decode/question</li>
</ul>
shaozk - neetcode

编码和解码字符串

链接</h1> https://neetcode.io/problems/string-encode-and-decode/question</li> </ul>

链接</h1>

https://neetcode.io/problems/string-encode-and-decode/question</li> </ul>
