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
- Ubuntu16.04LTS国内快速源
一.源文件位置 备份并替换/etc/apt/sources.list的源内容: 二.更改源文件内容 sudo vi /etc/apt/sources.list deb http://mirrors.a ...
- 关于Hibernate 的数据库配置
<hibernate-configuration> <session-factory name="mySessionFactory"> ...
- Google在KDD2013上关于CTR的一篇论文
最近在做CTR,刚好Google在KDD发了一篇文章,讲了他们的一些尝试,总结一下: 先是一些公式的符号说明: 一.优化算法 CTR中经常用Logistic regression进行训练,一个常用的L ...
- 对B+树与索引在MySQL中的认识
[TOC] 概述 本质:数据库维护某种数据结构以某种方式引用(指向)数据 索引取舍原则:索引的结构组织要尽量减少查找过程中磁盘I/O的存取次数 B树 满足的条件 d为大于1的一个正整数,称为B-Tre ...
- 7 Must Read Python Books
7 Must Read Python Books I started learning Python just two years ago. Coming from a C++ and Java ba ...
- 你们信不信一句Console.WriteLine就能让你的控制台程序失去响应
好久没更新博客了,今天是扒衣见君节,难得闲下来就来说说一个最近有趣的发现吧. 首先废话不多说,直接上代码吧 class Program { static void Main(string[] args ...
- Xamarin Error cannot find ‘aapt.exe’
Problem: solution: A workaround is to copy your files to the old directory. Just copy the aapt ...
- RS-232 vs. TTL Serial Communication(转载)
RS-232串口一度像现在的USB接口一样,是PC的标准接口,用来连接打印机.Modem和其他一些外设.后来逐渐被USB接口所取代,现在PC上已经看不到它的身影了.开发调试时如果用到串口,一般都是用U ...
- Bootstrap 简介
一.Bootstrap介绍 Bootstrap 是最受欢迎的 HTML.CSS 和 JS 框架,用于开发响应式布局.移动设备优先的 WEB 项目.本课时讲解 Bootstrap 的概念,并介绍 Boo ...
- NOIP2010关押罪犯[并查集|二分答案+二分图染色 | 种类并查集]
题目描述 S 城现有两座监狱,一共关押着N 名罪犯,编号分别为1~N.他们之间的关系自然也极不和谐.很多罪犯之间甚至积怨已久,如果客观条件具备则随时可能爆发冲突.我们用“怨气值”(一个正整数值)来表示 ...