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Commit 96c0036

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feat: add rust solution to lc problem: No.3407 (doocs#4730)
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‎solution/3400-3499/3407.Substring Matching Pattern/README.md‎

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<!-- solution:start -->
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### 方法一
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### 方法一:字符串匹配
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根据题目描述,`*` 可以被替换为零个或多个字符组成的任意字符序列,因此我们可以将模式字符串 $p$ 按照 `*` 分割成若干个子串,如果这些子串依次出现在字符串 $s$ 中且顺序不变,则说明 $p$ 可以变成 $s$ 的子字符串。
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因此,我们首先初始化一个指针 $i$ 指向字符串 $s$ 的起始位置,然后遍历模式字符串 $p$ 按照 `*` 分割得到的每个子串 $t,ドル在字符串 $s$ 中从位置 $i$ 开始查找子串 $t,ドル如果找到了,则将指针 $i$ 移动到子串 $t$ 的末尾位置继续查找下一个子串;如果找不到,则说明模式字符串 $p$ 不能变成字符串 $s$ 的子字符串,返回 $\text{false}$。如果所有子串都找到了,则返回 $\text{true}$。
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时间复杂度 $O(n \times m),ドル空间复杂度 $O(m)$。其中 $n$ 和 $m$ 分别是字符串 $s$ 和模式字符串 $p$ 的长度。
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<!-- tabs:start -->
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‎solution/3400-3499/3407.Substring Matching Pattern/README_EN.md‎

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<!-- solution:start -->
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### Solution 1
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### Solution 1: String Matching
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According to the problem description, `*` can be replaced by any sequence of zero or more characters, so we can split the pattern string $p$ by `*` into several substrings. If these substrings appear in order in the string $s,ドル then $p$ can become a substring of $s$.
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Therefore, we first initialize a pointer $i$ to the start of string $s,ドル then iterate over each substring $t$ obtained by splitting the pattern string $p$ by `*`. For each $t,ドル we search for it in $s$ starting from position $i$. If it is found, we move the pointer $i$ to the end of $t$ and continue searching for the next substring. If it is not found, it means the pattern string $p$ cannot become a substring of $s,ドル and we return $\text{false}$. If all substrings are found, we return $\text{true}$.
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The time complexity is $O(n \times m),ドル and the space complexity is $O(m),ドル where $n$ and $m$ are the lengths of strings $s$ and $p,ドル respectively.
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}
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```
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#### Rust
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```rust
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impl Solution {
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pub fn has_match(s: String, p: String) -> bool {
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let mut i = 0usize;
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for t in p.split('*') {
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if let Some(j) = s[i..].find(t) {
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i += j + t.len();
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} else {
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return false;
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}
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}
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true
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}
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}
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```
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<!-- tabs:end -->
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<!-- solution:end -->
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impl Solution {
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pub fn has_match(s: String, p: String) -> bool {
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let mut i = 0usize;
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for t in p.split('*') {
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if let Some(j) = s[i..].find(t) {
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i += j + t.len();
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} else {
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return false;
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}
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}
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true
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}
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}

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