说到回文数,大家可能会比较的陌生,但是在我们的日常生活中常会遇到这样的数字,只是你不知道它是回文数罢了. 例如:12321,这组数字就是回文数. 设n是一任意自然数.若将n的各位数字反向排列所得自然数n1与n相等,则称n为一回文数,这是大百度为我们的解释. 如果想更深入的了解,可以自行查找资料加深学习. 方法一: num = input("输入一个数") if num.isdigit(): num = str(num) for i in range(len(num)//2): if n
Given a string S, find the number of different non-empty palindromic subsequences in S, and return that number modulo 10^9 + 7. A subsequence of a string S is obtained by deleting 0 or more characters from S. A sequence is palindromic if it is equal
HDU 3294 Problem Description One day, sailormoon girls are so delighted that they intend to research about palindromic strings. Operation contains two steps:First step: girls will write a long string (only contains lower case) on the paper. For exa
原文地址: http://articles.leetcode.com/2011/11/longest-palindromic-substring-part-i.html 转载请注明出处:http://www.cnblogs.com/zhxshseu/p/4947609.html 问题描述:Given a string S, find the longest palindromic substring in S. 这道题目是一个经典的动态规划DP http://challenge.greplin.
Given a string s, find the longest palindromic substring in s. You may assume that the maximum length of s is 1000. Example 1: Input: "babad" Output: "bab" Note: "aba" is also a valid answer. Example 2: Input: "cbbd"
time O(n^2*k) space O(n^2) class Solution { public: int palindromePartition(string s, int K) { //分成两步:第一步递归求将下标[i,j]变为回文子串的最小代价cost(i,j); //cost(i,j)=cost(i+1,j-1)+(s[i]!=s[j]?1:0); //第二步:利用cost(i,j)递归求解将0~i分为k个回文子串的最小代价dp(i,k); //dp(i,k)=min{dp(j,k
本题来自 Project Euler 第4题:https://projecteuler.net/problem=4 # Project Euler: Problem 4: Largest palindrome product # A palindromic number reads the same both ways. # The largest palindrome made from the product # of two 2-digit numbers is 9009 = 91 × 9