Exercise 3: Check for Palindrome What do we have to achieve? This exercise focuses on checking if a given word is a palindrome. def is_it_palindrome(word): return word == word[::-1] # Test the function test_word = "level" print(f"Is '{test_word}' a palindrome? {is_it_palindrom...
If the linked list is not a palindrome, print "Not Palindrome". I am getting a segmentation fault after using the following functions. //The function for creating the nodes using array. struct node* create(head){ int n; printf("Enter the number of nodes in the linked list...
//Check if a given string is a rotation of a palindrome public class test7 { public static void main(String[] args) { String s = "Protijayi"; Set<Character> set = new HashSet<>(); for (int i = 0; i < s.length(); i++) { char ch = s.charAt(i)...
Python doctest Module | Document and Test CodeIn this tutorial, we will learn about the doctest module. It is a testing framework that helps us to document and test code simultaneously. This module allows us to document and test our code, which is essential to coding. By default, we can ...
https://leetcode.cn/problems/minimum-cost-to-convert-string-ii 这道题整体比较融合怪,正解是先预处理最短路矩阵,然后据此进行DP。但暴力切片判断某个子串能否转移,复杂度是不合理的,只是Python可以被放过。这个地方正确的做法是用字典树,因此这道题的码量还是比较大的。
classSolution:#@returna stringdeflongestPalindrome(self, s):iflen(s)==0:returns maxLen=1start=0foriinxrange(len(s)):ifi-maxLen >=1ands[i-maxLen-1:i+1]==s[i-maxLen-1:i+1][::-1]: start=i-maxLen-1maxLen+=2continueifi-maxLen >=0ands[i-maxLen:i+1]==s[i-maxLen:i+1...
(string) “232*+” Inputs: (string) str = “2*4*3+9*3+5″ Output: (string) “243**93*5++” The solution (in Python) is surprisingly short: def answer(normal): res = '' muls = normal.split('+') for mul in muls:
(n==0)return"^$";string ret="^";for(inti=0;i<n;i++)ret+="#"+s.substr(i,1);ret+="#$";returnret;}stringlongestPalindrome(string s){string T=preProcess(s);intn=T.length();int*P=newint[n];intC=0,R=0;for(inti=1;i<n-1;i++){inti_mirror=2*C-i;// equals to i' ...
解法1:水题,不过C++写起来不是很方便,用Python可能两行就搞定了。 // Solution 1: Trivial for Python, a little work for C++. 代码1 //Code 1 600 Non-negative Integers without Consecutive Ones // #600 不含连续1的非负整数 描述:给定一个数N,求所有不超过N,而且二进制表示中不含连续1的非负整数...
Code Llama 34b Python Non-Fine-Tuned Response: ### CodedeflongestPalindrome(s):# Fill this in.longest=""foriinrange(len(s)):forjinrange(len(s),i,-1):iflen(longest)>=j-i:breakelifs[i:j]==s[i:j][::-1]:longest=s[i:j]breakreturnlongestprint(longestPalindrome("abaxyzzyxf"))#...