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添加了问题“1807. 替换字符串中的括号内容”的代码(C++&Python)和题解
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| Original file line number | Diff line number | Diff line change |
|---|---|---|
| @@ -0,0 +1,35 @@ | ||
| /* | ||
| * @Author: LetMeFly | ||
| * @Date: 2023-01-12 22:15:05 | ||
| * @LastEditors: LetMeFly | ||
| * @LastEditTime: 2023-01-12 22:21:28 | ||
| */ | ||
| #ifdef _WIN32 | ||
| #include "_[1,2]toVector.h" | ||
| #endif | ||
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| class Solution { | ||
| public: | ||
| string evaluate(string s, vector<vector<string>>& knowledge) { | ||
| string ans; | ||
| unordered_map<string, string> ma; | ||
| for (auto& v : knowledge) { | ||
| ma[v[0]] = v[1]; | ||
| } | ||
| for (int i = 0; i < s.size(); i++) { | ||
| if (s[i] == '(') { | ||
| int to = i + 1; | ||
| while (s[to] != ')') { | ||
| to++; | ||
| } | ||
| string key = s.substr(i + 1, to - i - 1); | ||
| ans += ma.count(key) ? ma[key] : "?"; | ||
| i = to; // 循环结束后会有i++ | ||
| } | ||
| else { | ||
| ans += s[i]; | ||
| } | ||
| } | ||
| return ans; | ||
| } | ||
| }; |
This file contains bidirectional Unicode text that may be interpreted or compiled differently than what appears below. To review, open the file in an editor that reveals hidden Unicode characters.
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| Original file line number | Diff line number | Diff line change |
|---|---|---|
| @@ -0,0 +1,45 @@ | ||
| ''' | ||
| Author: LetMeFly | ||
| Date: 2023-01-12 22:23:47 | ||
| LastEditors: LetMeFly | ||
| LastEditTime: 2023-01-12 22:40:28 | ||
| ''' | ||
| from typing import List | ||
|
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| if False: | ||
| class Solution: | ||
| def evaluate(self, s: str, knowledge: List[List[str]]) -> str: | ||
| ma = {} | ||
| for v in knowledge: | ||
| ma[v[0]] = v[1] | ||
| ans = "" | ||
| for i in range(0, len(s)): # 不可，因为下面对i的更改无法作用到这里 | ||
| if (s[i] == '('): | ||
| to = i + 1 | ||
| while s[to] != ')': | ||
| to += 1 | ||
| ans += ma.get(s[i + 1 : to]) | ||
| i = to | ||
| else: | ||
| ans += s[i] | ||
| return ans | ||
|
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| # True | ||
| class Solution: | ||
| def evaluate(self, s: str, knowledge: List[List[str]]) -> str: | ||
| ma = {} | ||
| for v in knowledge: | ||
| ma[v[0]] = v[1] | ||
| ans = "" | ||
| i = 0 | ||
| while i < len(s): | ||
| if (s[i] == '('): | ||
| to = i + 1 | ||
| while s[to] != ')': | ||
| to += 1 | ||
| ans += ma.get(s[i + 1 : to], "?") | ||
| i = to | ||
| else: | ||
| ans += s[i] | ||
| i += 1 | ||
| return ans |
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| Original file line number | Diff line number | Diff line change |
|---|---|---|
| @@ -0,0 +1,163 @@ | ||
| --- | ||
| title: 1807.替换字符串中的括号内容 | ||
| date: 2023-01-12 22:22:24 | ||
| tags: [题解, LeetCode, 中等, 数组, 哈希表, 字符串, 字符串解析] | ||
| --- | ||
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| # 【LetMeFly】1807.替换字符串中的括号内容 | ||
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| 力扣题目链接：[https://leetcode.cn/problems/evaluate-the-bracket-pairs-of-a-string/](https://leetcode.cn/problems/evaluate-the-bracket-pairs-of-a-string/) | ||
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| <p>给你一个字符串 <code>s</code> ，它包含一些括号对，每个括号中包含一个 <strong>非空</strong> 的键。</p> | ||
|
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| <ul> | ||
| <li>比方说，字符串 <code>"(name)is(age)yearsold"</code> 中，有 <strong>两个</strong> 括号对，分别包含键 <code>"name"</code> 和 <code>"age"</code> 。</li> | ||
| </ul> | ||
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| <p>你知道许多键对应的值，这些关系由二维字符串数组 <code>knowledge</code> 表示，其中 <code>knowledge[i] = [key<sub>i</sub>, value<sub>i</sub>]</code> ，表示键 <code>key<sub>i</sub></code> 对应的值为 <code>value<sub>i</sub></code><sub> </sub>。</p> | ||
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| <p>你需要替换 <strong>所有</strong> 的括号对。当你替换一个括号对，且它包含的键为 <code>key<sub>i</sub></code> 时，你需要：</p> | ||
|
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| <ul> | ||
| <li>将 <code>key<sub>i</sub></code> 和括号用对应的值 <code>value<sub>i</sub></code> 替换。</li> | ||
| <li>如果从 <code>knowledge</code> 中无法得知某个键对应的值，你需要将 <code>key<sub>i</sub></code> 和括号用问号 <code>"?"</code> 替换（不需要引号）。</li> | ||
| </ul> | ||
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| <p><code>knowledge</code> 中每个键最多只会出现一次。<code>s</code> 中不会有嵌套的括号。</p> | ||
|
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| <p>请你返回替换 <strong>所有</strong> 括号对后的结果字符串。</p> | ||
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| <p> </p> | ||
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| <p><strong>示例 1：</strong></p> | ||
|
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| <pre> | ||
| <b>输入：</b>s = "(name)is(age)yearsold", knowledge = [["name","bob"],["age","two"]] | ||
| <b>输出：</b>"bobistwoyearsold" | ||
| <strong>解释：</strong> | ||
| 键 "name" 对应的值为 "bob" ，所以将 "(name)" 替换为 "bob" 。 | ||
| 键 "age" 对应的值为 "two" ，所以将 "(age)" 替换为 "two" 。 | ||
| </pre> | ||
|
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| <p><strong>示例 2：</strong></p> | ||
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| <pre> | ||
| <b>输入：</b>s = "hi(name)", knowledge = [["a","b"]] | ||
| <b>输出：</b>"hi?" | ||
| <b>解释：</b>由于不知道键 "name" 对应的值，所以用 "?" 替换 "(name)" 。 | ||
| </pre> | ||
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| <p><strong>示例 3：</strong></p> | ||
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| <pre> | ||
| <b>输入：</b>s = "(a)(a)(a)aaa", knowledge = [["a","yes"]] | ||
| <b>输出：</b>"yesyesyesaaa" | ||
| <b>解释：</b>相同的键在 s 中可能会出现多次。 | ||
| 键 "a" 对应的值为 "yes" ，所以将所有的 "(a)" 替换为 "yes" 。 | ||
| 注意，不在括号里的 "a" 不需要被替换。 | ||
| </pre> | ||
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| <p> </p> | ||
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| <p><strong>提示：</strong></p> | ||
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| <ul> | ||
| <li><code>1 <= s.length <= 10<sup>5</sup></code></li> | ||
| <li><code>0 <= knowledge.length <= 10<sup>5</sup></code></li> | ||
| <li><code>knowledge[i].length == 2</code></li> | ||
| <li><code>1 <= key<sub>i</sub>.length, value<sub>i</sub>.length <= 10</code></li> | ||
| <li><code>s</code> 只包含小写英文字母和圆括号 <code>'('</code> 和 <code>')'</code> 。</li> | ||
| <li><code>s</code> 中每一个左圆括号 <code>'('</code> 都有对应的右圆括号 <code>')'</code> 。</li> | ||
| <li><code>s</code> 中每对括号内的键都不会为空。</li> | ||
| <li><code>s</code> 中不会有嵌套括号对。</li> | ||
| <li><code>key<sub>i</sub></code> 和 <code>value<sub>i</sub></code> 只包含小写英文字母。</li> | ||
| <li><code>knowledge</code> 中的 <code>key<sub>i</sub></code> 不会重复。</li> | ||
| </ul> | ||
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| ## 方法一：哈希表 + 字符串解析 | ||
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| 这道题的关键就是字符串的解析，将字符串解析出来，然后将括号中的$key$替换成$value$，并且去掉括号即可。 | ||
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| 这就包含两个问题 | ||
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| 1. 如何将括号中的$key$解析出来 | ||
| 2. 如何将括号中的$key$替换为对应的$value$ | ||
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| **首先来解决问题一**：何将括号中的$key$解析出来 | ||
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| 我们用i遍历字符串$s$的下标，$i$的初始值是$0$，到$s.length$为止。 | ||
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| 期间，如果$s[i]$不为```'('```，那么我们就不断地将$s[i]$添加到答案字符串中去（这些都是不用解析的部分） | ||
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| 一旦遇到了```'('```，我们就用另外一个变量$to$，从$i + 1$开始往后累加，直到$s[to]$为```')'```为止。 | ||
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| 这样，我们就提取出了这对括号中间的$key$ | ||
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| **接着来解决问题二**：如何将括号中的$key$替换为对应的$value$ | ||
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| 这个很简单，开辟一个哈希表，首先遍历一遍$knowledge$数组，将$knowledge$中每一个“二元对”的第一个字符串当作$key$，第二个字符串当作$value$存入哈希表中。 | ||
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| 这样就能很方便地直到$key$是否在哈希表中，以及其在哈希表中的话，对应的$value$是什么 | ||
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| + 时间复杂度$O(len(s) + knowledge中字符数目之和)$ | ||
| + 空间复杂度$O(knowledge中字符数目之和)$ | ||
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| ### AC代码 | ||
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| #### C++ | ||
|
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| ```cpp | ||
| class Solution { | ||
| public: | ||
| string evaluate(string s, vector<vector<string>>& knowledge) { | ||
| string ans; | ||
| unordered_map<string, string> ma; | ||
| for (auto& v : knowledge) { | ||
| ma[v[0]] = v[1]; | ||
| } | ||
| for (int i = 0; i < s.size(); i++) { | ||
| if (s[i] == '(') { | ||
| int to = i + 1; | ||
| while (s[to] != ')') { | ||
| to++; | ||
| } | ||
| string key = s.substr(i + 1, to - i - 1); | ||
| ans += ma.count(key) ? ma[key] : "?"; | ||
| i = to; // 循环结束后会有i++ | ||
| } | ||
| else { | ||
| ans += s[i]; | ||
| } | ||
| } | ||
| return ans; | ||
| } | ||
| }; | ||
| ``` | ||
| #### Python | ||
| ```python | ||
| class Solution: | ||
| def evaluate(self, s: str, knowledge: List[List[str]]) -> str: | ||
| ma = {} | ||
| for v in knowledge: | ||
| ma[v[0]] = v[1] | ||
| ans = "" | ||
| i = 0 | ||
| while i < len(s): # 使用for i in range(s)的话，不易在下方修改i的值 | ||
| if (s[i] == '('): | ||
| to = i + 1 | ||
| while s[to] != ')': | ||
| to += 1 | ||
| ans += ma.get(s[i + 1 : to], "?") | ||
| i = to | ||
| else: | ||
| ans += s[i] | ||
| i += 1 | ||
| return ans | ||
| ``` | ||
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| > 同步发文于CSDN，原创不易，转载请附上[原文链接](https://leetcode.letmefly.xyz/2023/01/12/LeetCode%201807.%E6%9B%BF%E6%8D%A2%E5%AD%97%E7%AC%A6%E4%B8%B2%E4%B8%AD%E7%9A%84%E6%8B%AC%E5%8F%B7%E5%86%85%E5%AE%B9/)哦~ | ||
| > Tisfy：[https://letmefly.blog.csdn.net/article/details/128667712](https://letmefly.blog.csdn.net/article/details/128667712) |