回文串---Palindrome】的更多相关文章

POJ   3974 Description Andy the smart computer science student was attending an algorithms class when the professor asked the students a simple question, "Can you propose an efficient algorithm to find the length of the largest palindrome in a string…
Given a string s, partition s such that every substring of the partition is a palindrome. Return all possible palindrome partitioning of s. Example: Input: "aab" Output: [ ["aa","b"], ["a","a","b"…
[抄题]: 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. 给出 s = "aab",返回 [ ["aa", "b"], ["a", "a", "b"] ] [思维问题]: [一句话思路]: [输入量]:空: 正常情况:特大:特小:程序里处理到的特殊情况:异常情况(不合法不合理的输入): [画图]: [一刷]: 主函数中要记得新建parti…
131. 分割回文串 131. Palindrome Partitioning 题目描述 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 所有可能的分割方案. LeetCode131. Palindrome Partitioning中等 示例: 输入: "aab" 输出: [   ["aa","b"],   ["a","a","b"]] Java 实现 略…
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","…
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,…
题意:输入一个字符串,至少插入几个字符可以变成回文串(左右对称的字符串) 分析:f[x][y]代表x与y个字符间至少插入f[x][y]个字符可以变成回文串,可以利用动态规划的思想,求解 状态转化方程: f[x][y]=0  初始化 f[x][y]=f[x+1][y-1]     s[x]==s[y]时 f[x][y]=MIN ( f[x+1][y] , f[x][y-1] )+1    s[x] != s[y]时 代码展示 一开始没有将算的的数据存放到数组中,使得有些数据重复计算好多次,TLE…