There is a strange printer with the following two special requirements: The printer can only print a sequence of the same character each time. At each turn, the printer can print new characters starting from and ending at any places, and will cover t…

题目: There is a strange printer with the following two special requirements: The printer can only print a sequence of the same character each time. At each turn, the printer can print new characters starting from and ending at any places, and will cov…

There is a strange printer with the following two special requirements: The printer can only print a sequence of the same character each time. At each turn, the printer can print new characters starting from and ending at any places, and will cover t…

奇怪的打印机 有台奇怪的打印机有以下两个特殊要求: 打印机每次只能打印同一个字符序列. 每次可以在任意起始和结束位置打印新字符,并且会覆盖掉原来已有的字符. 给定一个只包含小写英文字母的字符串,你的任务是计算这个打印机打印它需要的最少次数. 示例 1: 输入: "aaabbb" 输出: 2 解释: 首先打印 "aaa" 然后打印 "bbb". 示例 2: 输入: "aba" 输出: 2 解释: 首先打印 "aaa&q…

There is a strange printer with the following two special requirements: The printer can only print a sequence of the same character each time. At each turn, the printer can print new characters starting from and ending at any places, and will cover t…

664. 奇怪的打印机 有台奇怪的打印机有以下两个特殊要求: 打印机每次只能打印同一个字符序列. 每次可以在任意起始和结束位置打印新字符,并且会覆盖掉原来已有的字符. 给定一个只包含小写英文字母的字符串,你的任务是计算这个打印机打印它需要的最少次数. 示例 1: 输入: "aaabbb" 输出: 2 解释: 首先打印 "aaa" 然后打印 "bbb". 示例 2: 输入: "aba" 输出: 2 解释: 首先打印 "…

There is a strange printer with the following two special requirements: The printer can only print a sequence of the same character each time. At each turn, the printer can print new characters starting from and ending at any places, and will cover t…

题意: 有一座电梯,其中楼层从1-n,每层都有一个数字k,当处于某一层时,只能往上走k层,或者下走k层.楼主在a层,问是否能到达第b层? 思路: 在起点时只能往上走和往下走两个选择,之后的每层都是这样,那么就类似于二叉树.每个节点就是对应的层,因为有可能碰到循环的层,比如1跳到3,3跳回1,这样使得无限循环,所以加个vis数组标记是否遍历过即可. #include <iostream> #include <cmath> #include <cstring> #inclu…

class Solution { public: int dp[100][100]; int dfs(const string &s, int i,int j) { if(i>j)return 0; if(dp[i][j])return dp[i][j]; dp[i][j]=dfs(s,i,j-1)+1; for(int k=i;k<j;++k) { if(s[k]==s[j])dp[i][j]=min(dp[i][j],dfs(s,i,k)+dfs(s,k+1,j-1)); } re…

一.这是个基础问题 Ref: Python之队列模拟算法(打印机问题)[首先研究这个问题作为开始] 任务队列 定义一个任务队列,来管理任务,而无需关心队列的”任务类型". # 自定义队列类 class Queue(object): def __init__(self): #初始化空队列 self.items = [] def isEmpty(self): #是否为空 return self.items == [] def enqueue(self, item): #入队,索引为0,即队尾在左侧…