[LintCode] 有效回文串】的更多相关文章

class Solution { public: /** * @param s A string * @return Whether the string is a valid palindrome */ bool isPalindrome(string& s) { // Write your code here , right = s.length() - ; while (left < right) { while (left < right && !isdigit…
题目: 有效回文串 给定一个字符串,判断其是否为一个回文串.只包含字母和数字,忽略大小写. 样例 "A man, a plan, a canal: Panama" 是一个回文. "race a car" 不是一个回文. 注意 你是否考虑过,字符串有可能是空字符串?这是面试过程中,面试官常常会问的问题. 在这个题目中,我们将空字符串判定为有效回文. 挑战 O(n) 时间复杂度,且不占用额外空间. 解题: 去除非有效字符后,整体是个回文串,两边查找,利用快速排序的思想,…
题目: 分割回文串 给定一个字符串s,将s分割成一些子串,使每个子串都是回文串. 返回s所有可能的回文串分割方案. 样例 给出 s = "aab",返回 [ ["aa","b"], ["a","a","b"] ] 解题: 这个题目不好搞啊,需要动态规划 在这里,没有根据动态规划,也解决了,貌似是暴力解决 从下标pos开始,找到下标i使得 pos到i内是回文字符串,再从i+1开始,找到下一…
108-分割回文串 II 给定一个字符串s,将s分割成一些子串,使每个子串都是回文. 返回s符合要求的的最少分割次数. 样例 比如,给出字符串s = "aab", 返回 1, 因为进行一次分割可以将字符串s分割成["aa","b"]这样两个回文子串 标签 动态规划 方法一(大体上没问题,但会在样例aaa....aaa上超时) 使用一维数组 dp[i] 记录s[i]...s[n-1]的最小切割数,状态转移方程为:dp[i] = min{dp[j]+…
Given a string which consists of lowercase or uppercase letters, find the length of the longest palindromes that can be built with those letters. This is case sensitive, for example "Aa" is not considered a palindrome here. Note: Assume the leng…
Given a string S, you are allowed to convert it to a palindrome by adding characters in front of it. Find and return the shortest palindrome you can find by performing this transformation. For example: Given "aacecaaa", return "aaacecaaa&qu…
Given a string s, partition s such that every substring of the partition is a palindrome. Return the minimum cuts needed for a palindrome partitioning of s. For example, given s = "aab", Return 1 since the palindrome partitioning ["aa"…
Given a string s, partition s such that every substring of the partition is a palindrome. Return all possible palindrome partitioning of s. For example, given s = "aab",Return [ ["aa","b"], ["a","a","…
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. 这道题让我们求最长回文子串,首先说下什么是回文串,就是正读反读都一样的字符串,比如 "bob", "level", &q…
考虑每个回文串,它一定是它中心字母的最长回文串两侧去掉同样数量的字符后的一个子串. 所以我们可以用manachar求出每一位的回文半径,放到哈希表里并标记出它的下一个子串. 最后拓扑排序递推就行了... 这道题丧心病狂卡哈希....wa了一屏... #include<iostream> #include<cstdio> #include<cstring> #include<algorithm> #include<queue> #include&l…