给定一个字符串，你的任务是计算这个字符串中有多少个回文子串。
具有不同开始位置或结束位置的子串，即使是由相同的字符组成，也会被视作不同的子串。
示例 1：
输入："abc"
输出：3
解释：三个回文子串: "a", "b", "c"
示例 2：
输入："aaa"
输出：6
解释：6个回文子串: "a", "a", "a", "aa", "aa", "aaa"
提示：
输入的字符串长度不会超过 1000 。
1 fn count_substrings(s: String) -> i32 { 2 let len = s.len(); 3 let temp_str = &s as &str; 4 let mut count = 0; 5 for gap in 1..=len { 6 for index in 0..len { 7 if len < index + gap { continue; } 8 let inner_str = &temp_str[index..index+gap]; 9 if is_palindromic(inner_str) { 10 count += 1; 11 } 12 } 13 } 14 count 15 } 16 17 18 fn is_palindromic(substring: &str) -> bool { 19 // let len = substring.len(); 20 // 21 // for i in 0..(len / 2) { 22 // let lhs = &substring[i..i + 1]; 23 // let rhs = &substring[len - i - 1..len - i]; 24 // if lhs.eq(rhs) { 25 // continue; 26 // } else { 27 // return false; 28 // } 29 // } 30 // //遍历完所有正确返回true 31 // true 32 33 let temp = &substring 34 .chars() 35 .rev() 36 .collect::<String>()[..]; 37 38 temp == substring 39 } 40 41 fn main() { 42 // let ret = is_palindromic("ab"); 43 // aaaaa == 15 44 // aaa == 6 45 // abc == 3 46 let s = String::from("aaaaa"); // 15 47 let ret = count_substrings(s); 48 println!("ret = {}", ret); 49 }
C++ Solution 超时
1 learn string operator 2 1. 截取子串 3 s.substr(pos, n) 截取s中从pos开始（包括0）的n个字符的子串，并返回 4 s.substr(pos) 截取s中从从pos开始（包括0）到末尾的所有字符的子串，并返回 5 2. 替换子串 6 s.replace(pos, n, s1) 用s1替换s中从pos开始（包括0）的n个字符的子串 7 3. 查找子串 8 s.find(s1) 查找s中第一次出现s1的位置，并返回（包括0） 9 s.rfind(s1) 查找s中最后次出现s1的位置，并返回（包括0） 10 s.find_first_of(s1) 查找在s1中任意一个字符在s中第一次出现的位置，并返回（包括0） 11 s.find_last_of(s1) 查找在s1中任意一个字符在s中最后一次出现的位置，并返回（包括0） 12 s.fin_first_not_of(s1) 查找s中第一个不属于s1中的字符的位置，并返回（包括0） 13 s.fin_last_not_of(s1) 查找s中最后一个不属于s1中的字符的位置，并返回（包括0） 14 * */ 15 16 17 18 #include <iostream> 19 #include <algorithm> 20 #include <string> 21 22 using namespace std; 23 bool is_palindromic(string s); 24 25 int countSubstrings(string s) { 26 int length = s.length(); 27 // cout << "length = " << length << endl; 28 int count = 0; 29 for (int gap = 1; gap <= length; gap++){ 30 for (int index = 0; index < length; index++){ 31 if (length < index + gap) { 32 continue; 33 } 34 string inner_str = s.substr(index, gap); 35 // cout << "inner_str = " << inner_str << endl; 36 if (is_palindromic(inner_str)) { 37 count += 1; 38 } 39 } 40 } 41 return count; 42 } 43 44 bool is_palindromic(string s) { 45 int length = s.length(); 46 for (int i = 0; i < length/2; i++){ 47 string lhs = s.substr(i,1); 48 string rhs = s.substr(length-i-1,1); 49 if (lhs == rhs) { 50 continue; 51 }else { 52 return false; 53 } 54 } 55 return true; 56 } 57 58 59 int main() { 60 string s = "aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa"; 61 62 int ret = countSubstrings(s); 63 // bool ret = is_palindromic(s); 64 cout << "ret = " << ret << endl; 65 return 0; 66 }
Python Soluton 超时
1 # 代码超时 2 # 3 # 4 # python string slice 5 # 6 # 切片操作（slice）可以从一个字符串中获取子字符串（字符串的一部分）。我们使用一对方括号、起始偏移量start、终止偏移量end 以及可选的步长step 来定义一个分片。 7 # 8 # 格式： [start:end:step] 9 # 10 # • [:] 提取从开头（默认位置0）到结尾（默认位置-1）的整个字符串 11 # • [start:] 从start 提取到结尾 12 # • [:end] 从开头提取到end - 1 13 # • [start:end] 从start 提取到end - 1 14 # • [start:end:step] 从start 提取到end - 1，每step 个字符提取一个 15 # • 左侧第一个字符的位置/偏移量为0，右侧最后一个字符的位置/偏移量为-1 16 # 17 # 18 # 19 # 几个特别的examples 如下： 20 # 21 # 提取最后N个字符： 22 # 23 # >>> letter = 'abcdefghijklmnopqrstuvwxyz' 24 # >>> letter[-3:] 25 # 'xyz' 26 # 27 # 从开头到结尾，step为N： 28 # 29 # >>> letter[::5] 30 # 'afkpuz' 31 # 32 # 将字符串倒转(reverse)， 通过设置步长为负数： 33 # 34 # >>> letter[::-1] 35 # 'zyxwvutsrqponmlkjihgfedcba' 36 37 38 39 def is_palindromic(s: str) -> bool: 40 length: int = len(s) 41 for i in range(length // 2): 42 lhs = s[i:i + 1] 43 rhs = s[length - i - 1:length - i] 44 if lhs == rhs: 45 continue 46 else: 47 return False 48 return True 49 50 51 def countSubstrings(s: str) -> int: 52 # def is_palindromic(s: str) -> bool: 53 # _length: int = len(s) 54 # for i in range(_length // 2): 55 # lhs = s[i:i + 1] 56 # rhs = s[_length - i - 1:_length - i] 57 # if lhs == rhs: 58 # continue 59 # else: 60 # return False 61 # return True 62 63 length: int = len(s) 64 count = 0 65 for gap in range(length): 66 new_gap = gap + 1 67 for index in range(length): 68 if length < index + new_gap: 69 continue 70 inner_str = s[index:index + new_gap] 71 if is_palindromic(inner_str): 72 count += 1 73 74 return count 75 76 77 if __name__ == "__main__": 78 # print("is_palindromic {}".format(is_palindromic("abc"))) 79 80 print("ret = {}".format(countSubstrings("aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa")))
Github Link: https://github.com/DaviRain-Su/leetcode_solution/tree/master/palindromic_substrings