目录

28. 找出字符串中第一个匹配项的下标 Find-the-index-of-the-first-occurrence-in-a-string  🌟🌟

29. 两数相除 Divide Two Integers  🌟🌟

30. 串联所有单词的子串 Substring-with-concatenation-of-all-words  🌟🌟🌟

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28. 找出字符串中第一个匹配项的下标 Find-the-index-of-the-first-occurrence-in-a-string

给你两个字符串 haystack 和 needle ，请你在 haystack 字符串中找出 needle 字符串出现的第一个位置（下标从 0 开始）。如果不存在，则返回  -1 。

说明：实现 ​strStr()​ 函数。对于本题而言，当 needle 是空字符串时我们应当返回 0 。这与 C 语言的 ​strStr()​ 以及 Java 的​ indexOf()​ 定义相符。

当 needle 是空字符串时，我们应当返回什么值呢？这是一个在面试中很好的问题。

示例 1：

输入：haystack = "hello", needle = "ll"
输出：2

示例 2：

输入：haystack = "aaaaa", needle = "bba"
输出：-1

提示：

• 1 <= haystack.length, needle.length <= 10^4
• haystack 和 needle 仅由小写英文字符组成

代码1：

fn str_str(haystack: String, needle: String) -> i32 {let n = haystack.len();let m = needle.len();if m == 0 {return 0;}if n < m {return -1;}for i in 0..=n-m {if haystack[i..i+m] == needle {return i as i32;}}return -1;
}fn main() {let haystack = "hello".to_string();let needle = "ll".to_string();println!("{}", str_str(haystack, needle));let haystack = "aaaaa".to_string();let needle = "bba".to_string();println!("{}", str_str(haystack, needle));
}

输出：

2
-1

代码2：

fn str_str(haystack: String, needle: String) -> i32 {let mut i = 0_usize;loop {let mut j = 0_usize;loop {if j == needle.len() {return i as i32;}if i + j == haystack.len() {return -1;}if needle.as_bytes()[j] != haystack.as_bytes()[i+j] {break;}j += 1;}i += 1;}
}fn main() {let haystack = "hello".to_string();let needle = "ll".to_string();println!("{}", str_str(haystack, needle));let haystack = "aaaaa".to_string();let needle = "bba".to_string();println!("{}", str_str(haystack, needle));
}

另： Rust语言有现成的字符串方法 haystack.find(&needle)

fn main() {let haystack = "hello".to_string();let needle = "ll".to_string();println!("{:?}", haystack.find(&needle));let haystack = "aaaaa".to_string();let needle = "bba".to_string();println!("{:?}", haystack.find(&needle));
}

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29. 两数相除 Divide Two Integers

给定两个整数，被除数 dividend 和除数 divisor。将两数相除，要求不使用乘法、除法和 mod 运算符。

返回被除数 dividend 除以除数 divisor 得到的商。

整数除法的结果应当截去（truncate）其小数部分，例如：truncate(8.345) = 8 以及 truncate(-2.7335) = -2

示例 1:

输入: dividend = 10, divisor = 3
输出: 3
解释: 10/3 = truncate(3.33333..) = truncate(3) = 3

示例 2:

输入: dividend = 7, divisor = -3
输出: -2
解释: 7/-3 = truncate(-2.33333..) = -2

提示：

• 被除数和除数均为 32 位有符号整数。
• 除数不为 0。
• 假设我们的环境只能存储 32 位有符号整数，其数值范围是 [−2^31,  2^31 − 1]。本题中，如果除法结果溢出，则返回 2^31 − 1。

代码：

pub fn divide(dividend: i32, divisor: i32) -> i32 {// 处理特殊情况if dividend == std::i32::MIN && divisor == -1 {return std::i32::MAX;}if divisor == 1 {return dividend;}if divisor == -1 {return -dividend;}// 处理符号let mut res = 0;let mut sign = 1;if (dividend > 0 && divisor < 0) || (dividend < 0 && divisor > 0) {sign = -1;}let mut a = abs(dividend);let b = abs(divisor);// 计算商while a >= b {let (mut temp, mut tb) = (1, b);while a >= (tb << 1) {tb <<= 1;temp <<= 1;}res += temp;a -= tb;}res * sign
}fn abs(x: i32) -> i32 {if x < 0 {-x} else {x}
}fn main() {println!("{}", divide(10, 3));println!("{}", divide(7, -3));
}

输出：

3
-2

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30. 串联所有单词的子串 Substring-with-concatenation-of-all-words

给定一个字符串 s 和一些 长度相同 的单词 words 。找出 s 中恰好可以由 words 中所有单词串联形成的子串的起始位置。

注意子串要与 words 中的单词完全匹配，中间不能有其他字符 ，但不需要考虑 words 中单词串联的顺序。

示例 1：

输入：s = "barfoothefoobarman", words = ["foo","bar"]
输出：[0,9]解释：
从索引 0 和 9 开始的子串分别是 "barfoo" 和 "foobar" 。
输出的顺序不重要, [9,0] 也是有效答案。

示例 2：

输入：s = "wordgoodgoodgoodbestword", words = ["word","good","best","word"]
输出：[]

示例 3：

输入：s = "barfoofoobarthefoobarman", words = ["bar","foo","the"]
输出：[6,9,12]

提示：

• 1 <= s.length <= 10^4
• s 由小写英文字母组成
• 1 <= words.length <= 5000
• 1 <= words[i].length <= 30
• words[i] 由小写英文字母组成

代码1：  暴力枚举

pub fn find_substring(s: String, words: Vec<String>) -> Vec<i32> {let n = s.len();let m = words.len();if n == 0 || m == 0 {return Vec::new();}let word_len = words[0].len();let mut ans = Vec::new();for i in 0..=n - m * word_len {let mut j = 0;let mut used = vec![false; m];while j < m {let word = &s[i + j * word_len..i + j * word_len + word_len];let mut k = 0;while k < m {if !used[k] && word == &words[k] {used[k] = true;break;}k += 1;}if k == m {break;}j += 1;}if j == m {ans.push(i as i32);}}ans
}fn main() {let s = String::from("barfoothefoobarman");let words = vec![String::from("foo"), String::from("bar")];println!("{:?}", find_substring(s, words));let s = String::from("wordgoodgoodgoodbestword");let words = vec![String::from("word"),String::from("good"),String::from("best"),String::from("word"),];println!("{:?}", find_substring(s, words));let s = String::from("barfoofoobarthefoobarman");let words = vec![String::from("bar"), String::from("foo"), String::from("the")];println!("{:?}", find_substring(s, words));
}

输出：

[0, 9]
[]
[6, 9, 12]

代码2： 滑动窗口

use std::collections::HashMap;pub fn find_substring(s: String, words: Vec<String>) -> Vec<i32> {let n = s.len();let m = words.len();if n == 0 || m == 0 {return vec![];}let word_len = words[0].len();let mut cnt = HashMap::new();for word in words {*cnt.entry(word).or_insert(0) += 1;}let mut ans = Vec::new();for i in 0..word_len {let mut left = i;let mut right = i;let mut window = HashMap::new();while right + word_len <= n {let word = &s[right..right + word_len];right += word_len;if *cnt.get(word).unwrap_or(&0) == 0 {left = right;window.clear();} else {*window.entry(word.to_string()).or_insert(0) += 1;while *window.get(word).unwrap_or(&0) > *cnt.get(word).unwrap_or(&0) {let d_word = &s[left..left + word_len];left += word_len;*window.entry(d_word.to_string()).or_insert(0) -= 1;}if right - left == word_len * m {ans.push(left as i32);}}}}ans
}fn main() {let s = String::from("barfoothefoobarman");let words = vec![String::from("foo"), String::from("bar")];println!("{:?}", find_substring(s, words));let s = String::from("wordgoodgoodgoodbestword");let words = vec![String::from("word"),String::from("good"),String::from("best"),String::from("word"),];println!("{:?}", find_substring(s, words));let s = String::from("barfoofoobarthefoobarman");let words = vec![String::from("bar"), String::from("foo"), String::from("the")];println!("{:?}", find_substring(s, words));
}

代码3：滑动窗口

use std::collections::HashMap;pub fn find_substring(s: String, words: Vec<String>) -> Vec<i32> {let n = s.len();let m = words.len();if n == 0 || m == 0 {return vec![];}let word_len = words[0].len();let mut cnt = HashMap::new();for word in &words {*cnt.entry(word.to_string()).or_insert(0) += 1;}let mut ans = Vec::new();for i in 0..word_len {let mut left = i;let mut right = i;let mut window = HashMap::new();let mut count = 0;while right + word_len <= n {let word = &s[right..right + word_len];right += word_len;if cnt.get(word).cloned().unwrap_or(0) == 0 {left = right;window.clear();count = 0;} else {*window.entry(word.to_string()).or_insert(0) += 1;count += 1;while window.get(word).cloned().unwrap_or(0) > cnt.get(word).cloned().unwrap_or(0) {let d_word = &s[left..left + word_len];left += word_len;*window.entry(d_word.to_string()).or_insert(0) -= 1;count -= 1;}if count == m {ans.push(left as i32);}}}}ans
}fn main() {let s = String::from("barfoothefoobarman");let words = vec![String::from("foo"), String::from("bar")];println!("{:?}", find_substring(s, words));let s = String::from("wordgoodgoodgoodbestword");let words = vec![String::from("word"),String::from("good"),String::from("best"),String::from("word"),];println!("{:?}", find_substring(s, words));let s = String::from("barfoofoobarthefoobarman");let words = vec![String::from("bar"), String::from("foo"), String::from("the")];println!("{:?}", find_substring(s, words));
}

代码4： 滑动窗口

use std::collections::HashMap;pub fn find_substring(s: String, words: Vec<String>) -> Vec<i32> {let n = s.len();let m = words.len();if n == 0 || m == 0 {return vec![];}let word_len = words[0].len();let mut cnt = HashMap::new();for word in &words {*cnt.entry(word.to_string()).or_insert(0) += 1;}let mut ans = Vec::new();for i in 0..word_len {let mut left = i;let mut right = i;let mut window = HashMap::new();let mut count = 0;while right + word_len <= n {let word = &s[right..right + word_len];right += word_len;if cnt.get(word).cloned().unwrap_or(0) == 0 {left = right;window.clear();count = 0;} else {*window.entry(word.to_string()).or_insert(0) += 1;count += 1;while window.get(word).cloned().unwrap_or(0) > cnt.get(word).cloned().unwrap_or(0) {let d_word = &s[left..left + word_len];left += word_len;*window.entry(d_word.to_string()).or_insert(0) -= 1;count -= 1;}if count == m {ans.push(left as i32);}}if right - left == word_len * (m + 1) {let d_word = &s[left..left + word_len];left += word_len;*window.entry(d_word.to_string()).or_insert(0) -= 1;count -= 1;}}}ans
}fn main() {let s = String::from("barfoothefoobarman");let words = vec![String::from("foo"), String::from("bar")];println!("{:?}", find_substring(s, words));let s = String::from("wordgoodgoodgoodbestword");let words = vec![String::from("word"),String::from("good"),String::from("best"),String::from("word"),];println!("{:?}", find_substring(s, words));let s = String::from("barfoofoobarthefoobarman");let words = vec![String::from("bar"), String::from("foo"), String::from("the")];println!("{:?}", find_substring(s, words));
}

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