Have you consider that the string might be empty? This is a good question to ask during an interview. For the purpose of this problem, we define empty string as valid palindrome. 1.【基础知识】 回文字符串是一个正读和反读都一样的字符串,比方“level”或者“noon”等等就是回文串。 2.【屌...
https://leetcode.com/problems/valid-palindrome/ 题目: Given a string, determine if it is a palindrome, considering only alphanumeric characters and ignoring cases. For example, "A man, a plan, a canal: Panama"is a palindrome. "race a car"isnota palindrome. Note: Have you consider that ...
func validPalindrome(s string) bool { // 定义左指针 l ,初始化为 0 l := 0 // 定义右指针 r ,初始化为 s.length - 1 r := len(s) - 1 // 当还有字符需要比较时,继续处理 for l < r { // 如果 s[l] 和 s[r] 不相等,则需要删除字符 if s[l] != s[r] { // 如果删除 左指...
For the purpose of this problem, we define empty string as valid palindrome. 这道题考查的是指定字符的回文数字的判断。 建议和leetcode 680. Valid Palindrome II 去除一个字符的回文字符串判断 + 双指针 一起学习 代码如下: public class Solution { //判断是否是回文数:这个题只考虑小写字符和数字,别的...
建议和leetcode 680. Valid Palindrome II 去除一个字符的回文字符串判断 + 双指针 一起学习 代码如下: #include <iostream> #include <vector> #include <map> #include <set> #include <queue> #include <stack> #include <string> #include <climits> ...
[LintCode/LeetCode] Longest Palindrome Substring Problem Given a string S, find the longest palindromic substring in S. You may assume that the maximum length of S is 1000, and there exists one unique longest palindromic substring. Example...
Input: s = "leetcode" Output: 5 Explanation: Inserting 5 characters the string becomes "leetcodocteel". Constraints: 1 <= s.length <= 500 s consists of lowercase English letters. 题目描述: 给你一个字符串 s ,每一次操作你都可以在字符串的任意位置插入任意字符。
1147. 段式回文 - 你会得到一个字符串 text 。你应该把它分成 k 个子字符串 (subtext1, subtext2,…, subtextk) ,要求满足: * subtexti 是 非空 字符串 * 所有子字符串的连接等于 text ( 即subtext1 + subtext2 + ... + subtextk == text ) * 对于所有 i 的有效值(
Some hints: Could negative integers be palindromes? (ie, -1) If you are thinking of converting the integer to string, note the restriction of using extra space. You could also try reversing an integer. However, if you have solved the problem "Reverse Integer", you know that the reversed ...
class Solution { int seg = 0; public int longestDecomposition(String text) { find(text); //最后一段统一加1 return seg+1; } private void find(String text) { //如果最后刚好为空,全部匹配上,先减1,因为后面会统一加1 if("".equals(text)) { seg--; return; } int l = 0; int r = ...