Palindrome Index
传送门: Palindrome Index
Problem Statement
You are given a string of lower case letters. Your task is to figure out the index of the character on whose removal it will make the string a palindrome. There will always be a valid solution.
In case the string is already a palindrome, then -1
is also a valid answer along with possible indices.
Input Format
The first line contains T, i.e. the number of test cases.
T lines follow, each containing a string.
Output Format
Print the position (0 index) of the letter by removing which the string turns into a palindrome. For a string, such as
bcbc
we can remove b at index 0 or c at index 3. Both answers are accepted.
Constraints
1≤T≤20
1≤ length of string ≤100005
All characters are Latin lower case indexed.
Sample Input
3
aaab
baa
aaa
Sample Output
3
0
-1
Explanation
In the given input, T = 3,
- For input aaab, we can see that removing b from the string makes the string a palindrome, hence the position 3.
- For input baa, removing b from the string makes the string palindrome, hence the position 0.
- As the string aaa is already a palindrome, you can output 0, 1 or 2 as removal of any of the characters still maintains the palindrome property. Or you can print -1 as this is already a palindrome.
读题时需注意:
题目中先说 “There will always be a valid solution. ”,然后才说“In case the string is already a palindrome, then -1
is also a valid answer along with possible indices.”。注意体会这句话,我们首先应注意到,即使输入的字符串S是个回文串,也可以删除某个字母使其仍为回文串。如果|S|为奇数,则删除中间那个字母,结果串仍为回文串。如果|S|为偶数则删除中间两个相等字符中的任一个,结果串也回文。
完全暴力的解法:
枚举要删除的字母,检查结果串是否回文。复杂度O(N^2)。
#include<bits/stdc++.h>
using namespace std;
const int MAX_N=1e5+;
char s[MAX_N];
int len;
int opp(int j, int x){
if(x==){
return len+-j;
}
if(j<x){
return len-j<x? len-j: len-j+;
}
else{
return len+-j;
}
}
bool ok(int x){
int tmp=x?(len-)>>:len>>;
for(int i=, j=; i<tmp; i++, j++){
if(j==x){
j++;
}
if(s[j]!=s[opp(j, x)]){
return false;
}
}
return true;
}
int main(){
int T;
scanf("%d", &T);
while(T--){
scanf("%s", s+);
len=strlen(s+);
for(int i=; i<=len; i++){
if(ok(i)){
printf("%d\n", i-);
break;
}
}
}
return ;
}
只是这解法过于暴力,TLE。
下面就要引入这道题给我的最大启示了:
寻找有助于简化问题的必要条件
考虑一下上面的单纯暴力算法有那些冗余计算。
首先必须指出一个问题:优化算法的途径是充分考虑问题的特殊性。
其次要注意到:题目要求的是存在性判别,上面的算法枚举被删除字符的位置是无可厚非的。
接着考虑一下使上面的算法达到最坏情况的数据:
aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaab
在这种情况下,上述算法必须枚举到最后一个字符才能确定答案。
我们不难发现一个问题
Palindrome Index的更多相关文章
- [hackerrank]Palindrome Index
简单题. #include <iostream> #include <string> using namespace std; int main() { int T; cin ...
- LintCode Palindrome Partitioning II
Given a string s, cut s into some substrings such that every substring is a palindrome. Return the m ...
- LeetCode 214 Shortest Palindrome
214-Shortest Palindrome Given a string S, you are allowed to convert it to a palindrome by adding ch ...
- 【Leetcode】【Medium】Palindrome Partitioning
Given a string s, partition s such that every substring of the partition is a palindrome. Return all ...
- Leetcode | Palindrome
Valid Palindrome Given a string, determine if it is a palindrome, considering only alphanumeric char ...
- POJ2402/UVA 12050 Palindrome Numbers 数学思维
A palindrome is a word, number, or phrase that reads the same forwards as backwards. For example,the ...
- [LeetCode#266] Palindrome Permutation
Problem: Given a string, determine if a permutation of the string could form a palindrome. For examp ...
- Palindrome Pairs 解答
Question Given a list of unique words, find all pairs of distinct indices (i, j) in the given list, ...
- Reverse Integer - Palindrome Number - 简单模拟
第一个题目是将整数进行反转,这个题实现反转并不难,主要关键点在于如何进行溢出判断.溢出判断再上一篇字符串转整数中已有介绍,本题采用其中的第三种方法,将数字转为字符串,使用字符串比较大小的方法进行比较. ...
随机推荐
- mysql unrecognized service问题解决
在centos下用yum install mysql 安装完后,却发现用service mysqld start无法开启 出现mysqld:unrecognized service,网上别人说用/ ...
- ubuntu系统安装软件方法
ubuntu系统安装软件方法 ubuntu下安装软件有三种方式,分别为在线安装apt-get方式,软件商店安装方式和 1. 软件商店安装方式 这种方式对经常使用windows系统的同学来说最为简单,因 ...
- Consul Windows 安装
下载文件https://www.consul.io/downloads.html, 解压完毕后只有一个consul文件 consul 启动一个 Agent consul agent -server - ...
- android 开启本地相册选择图片并返回显示
.java package com.jerry.crop; import java.io.File; import android.app.Activity; import android.conte ...
- jackson官方快速入门文档
官方地址: http://jackson.codehaus.org/ http://wiki.fasterxml.com/JacksonInFiveMinutes http://wiki.faster ...
- Oracle数据库穿越防火墙访问
原因 Oracle listener 只起一个中介作用,当客户连接它时,它根据配置寻找到相应的数据库实例进程,然后spawned一个新的数据库连接,这个连接端口由listener传递给客户机,此后客户 ...
- [django]l利用xlrd实现xls文件导入数据
代码: #coding:utf-8 import os os.environ.setdefault("DJANGO_SETTINGS_MODULE", "www.sett ...
- 【小白的CFD之旅】03 老蓝
第一次见到老蓝,小白都不太敢相信,对面那不修边幅的糟老头子会是自己要找的导师.嘴里叼着烟,牙都掉了好几颗,穿着还算整齐,这是小白对老蓝的第一印象,这印象并不太好,尤其是在小白发誓认真度过研究生三年时光 ...
- Serial Communication Protocol Design Hints And Reference
前面转载的几篇文章详细介绍了UART.RS-232和RS-485的相关内容,可以知道,串口通信的双方在硬件层面需要约定如波特率.数据位.校验位和停止位等属性,才可以正常收发数据.实际项目中使用串口通信 ...
- UVA - 10375 Choose and divide[唯一分解定理]
UVA - 10375 Choose and divide Choose and divide Time Limit: 1000MS Memory Limit: 65536K Total Subm ...