palindrome专题

ural 1297. Palindrome dp

1297. Palindrome Time limit: 1.0 second Memory limit: 64 MB The “U.S. Robots” HQ has just received a rather alarming anonymous letter. It states that the agent from the competing «Robots Unli

【UVA】10739 - String to Palindrome(动态规划)

比较水的动态规划 dp[i][j] 将原串 i ~ j 之内的字符转化为回文字符所需要的最小操作次数 其中删除操作和添加操作本质上是一样的。 三个状态转移方程: dp[i][j] = min(dp[i][j] ,dp[i + 1][j]); dp[i][j] = min(dp[i][j] ,dp[i + 1][j - 1]); dp[i][j] = min(dp[i][j] ,dp[

leetCode#125. Valid Palindrome

Description 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

[LeetCode] 409. Longest Palindrome

题:https://leetcode.com/problems/longest-palindrome/description/ 题目 Given a string which consists of lowercase or uppercase letters, find the length of the longest palindromes that can be built with

[LeetCode] 234. Palindrome Linked List

题:https://leetcode.com/problems/palindrome-linked-list/description/ 题目 Given a singly linked list, determine if it is a palindrome. Example 1: Input: 1->2Output: false Example 2: Input: 1->2->

Count Palindrome in String

string 有多少palindrome substring。exp: 'aba' 返回4 , 'abba' 返回6 public class CountPalindrome {public int countPalindrome(String str){if(str == null || str.length() == 0) return 0;int count = 0;for(int

leetcode 刷题之路 28 Palindrome Number

Determine whether an integer is a palindrome. Do this without extra space. 判断一个数字是否是回文数字。 测试通过程序: class Solution {public:bool isPalindrome(int x){int revX=0;int xTemp=x;while (xTemp > 0){int tem

【LeetCode】Palindrome Partitioning III

leetcode DP 深搜 回文串 1 Palindrome Partitioning 问题来源:Palindrome Partitioning 该问题简单来说就是给定一个字符串,将字符串分成多个部分,满足每一部分都是回文串,请输出所有可能的情况。        该问题的难度比较大,很可能第一次遇到没有思路,这很正常。下面我们一点点分析,逐步理清思路。先不考虑所有的情况,针对一个

Codeforces Beta Round #7--D. Palindrome Degree(Manacer)

题目:http://blog.csdn.net/winddreams/article/details/44218961 求出每个点为中心的最长字符串,判断该串是不是从开头的回文串。   #include <cstdio>#include <cstring>#include <algorithm>using namespace std ;int p[12000000] , dp[600

Codeforces Beta Round #7--D. Palindrome Degree(hash)

D. Palindrome Degree time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output String s of length n is called k-palindrome, if it

Codeforces Round #311 (Div. 2) E. Ann and Half-Palindrome (DP+字典树)

题目地址:传送门 先用dp求出所有的符合要求的半回文串,标记出来。然后构造字典树。然后再dfs一遍求出所有节点的子树和,最后搜一遍就能找出第k个来了。 代码如下: #include <iostream>#include <string.h>#include <math.h>#include <queue>#include <algorithm>#include <stdlib.h>

lightoj 1044 Palindrome Partitioning(dp)

题意:给定字符串S,问可以划分的最小回文串数量 思路:定义dp[i]为以i开头的字符串中回文串的最小划分数. dp[i] = min(dp[j] + 1 | i <= j < n && S[i...j]是回文串) 边界,dp[i] = n-i+1. /************************************************ Author: fisty* Crea

**Palindrome Partitioning

题目: https://leetcode.com/problems/palindrome-partitioning/ Given a string s, partition s such that every substring of the partition is a palindrome. Return all possible palindrome partitioning

**Palindrome Partitioning II

题目: 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", R

Palindrome Number问题及解法

问题描述: Determine whether an integer is a palindrome. Do this without extra space. 求解一个数是不是回文数。 问题求解思路: 1.负数不是回文数 2.回文数正着读和反着读一样,故可将该数反转与原数比较大小 我的代码如下: class Solution {public:bool isPali

Largest Palindrome Product问题及解法

问题描述: Find the largest palindrome made from the product of two n-digit numbers. Since the result could be very large, you should return the largest palindrome mod 1337. 示例: Input: 2 Output: 987 E

Valid Palindrome II问题及解法

问题描述: Given a non-empty string s, you may delete at most one character. Judge whether you can make it a palindrome. 示例: Input: "aba"Output: True Input: "abca"Output: TrueExplanation: You c

Easy 3 Palindrome Number(9)

Description Determine whether an integer is a palindrome. Do this without extra space. Solution 判断一个整型数是否为回文。如果将integer全翻转,可能会导致整数溢出。但这并不影响判断,因为若是回文,则翻转必不会溢出。 class Solution {public:bool isPalind

HDU-1513 Palindrome LCS+滚动数组

/*http://acm.hdu.edu.cn/showproblem.php?pid=1513将原字符串倒置,然后与原字符串求最长公共子序列,答案就是len-dp[len][len]。*/#include "stdio.h"#include "string.h"const int maxn = 510;int n;char str1[maxn],str2[maxn];int d

【LeetCode最详尽解答】125-验证回文串 Valid-Palindrome

欢迎收藏Star我的Machine Learning Blog:https://github.com/purepisces/Wenqing-Machine_Learning_Blog。如果收藏star, 有问题可以随时与我交流, 谢谢大家! 链接: 125-验证回文串 直觉 这个问题需要使用一些内置函数,比如 s[l].isalnum() 和 s[l].lower()。基本要求简单明了。

动态规划+滚动数组 -- POJ 1159 Palindrome

给一字符串,问最少加几个字符可以让它成为回文串, 比如 Ab3bd 最少需要两个字符可以成为回文串 dAb3bAd 思路: 动态规划 DP[i][j] 意味着从 i 到 j 这段字符变为回文串最少要几个字符,枚举子串长。 if str[i] == str[j]: DP[i][j] = DP[i + 1][j - 1] else: DP[i][j] = min( DP[i

杭电1513(Palindrome)

点击打开杭电1513 Problem Description A palindrome is a symmetrical string, that is, a string read identically from left to right as well as from right to left. You are to write a program which, given a

【位运算】【前缀和】个人练习-Leetcode-1177. Can Make Palindrome from Substring

题目链接:https://leetcode.cn/problems/can-make-palindrome-from-substring/description/ 题目大意:给出一个字符串s,每次query给出l, r, k,要求判断子串s[l:r+1]在经过k次操作后是否能变为回文串。一次操作可以将子串内的一个字符变为任意一个其他字符。并且子串顺序可以任意改变。 思路:因为有很多query,

leetcode No9. Palindrome Number

Question: Determine whether an integer is a palindrome. Do this without extra space. 判断一个整数是否为回文数,回文数是把整数n的各位数字反向排列所得数与n相等。 Algorithm: 把n的各位数字反向排列得到的数和n相等即是回文数 Submitted Code: class Solutio

UVA - 11151 Longest Palindrome

题意:以为是求最长回文串,结果用Manacher 一直错,原来题目是求最长回文子串,汗!!那么就很好处理了,跟之前的一样,当str[l]==str[r]的时候,+1,否则在dp(l+1,r),dp(l,r-1)找最大的 #include <iostream>#include <cstdio>#include <cstring>#include <algorithm>using na

UVA - 10453 Make Palindrome

题意:跟之前删除,添加,替换转化成回文串一个意思,还是从后往前枚举,最后根据dp[i+1][j],dp[i][j-1]的大小,我们就可以判断是添加在哪一边了,如果dp[i+1][j]大的话,那么果断是复制s[l]在最右边,因为不相等的话我们肯定要添加一个,当然要找最小的啦,另一种情况就复制s[r]在最左边,打印的时候注意如果有中间的字符的时候要记得打印就行了 #include <iostream