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

  1. [hackerrank]Palindrome Index

    简单题. #include <iostream> #include <string> using namespace std; int main() { int T; cin ...

  2. LintCode Palindrome Partitioning II

    Given a string s, cut s into some substrings such that every substring is a palindrome. Return the m ...

  3. LeetCode 214 Shortest Palindrome

    214-Shortest Palindrome Given a string S, you are allowed to convert it to a palindrome by adding ch ...

  4. 【Leetcode】【Medium】Palindrome Partitioning

    Given a string s, partition s such that every substring of the partition is a palindrome. Return all ...

  5. Leetcode | Palindrome

    Valid Palindrome Given a string, determine if it is a palindrome, considering only alphanumeric char ...

  6. 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 ...

  7. [LeetCode#266] Palindrome Permutation

    Problem: Given a string, determine if a permutation of the string could form a palindrome. For examp ...

  8. Palindrome Pairs 解答

    Question Given a list of unique words, find all pairs of distinct indices (i, j) in the given list, ...

  9. 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

  1. 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 ...

  2. 关于Hibernate 的数据库配置

    <hibernate-configuration>    <session-factory name="mySessionFactory">        ...

  3. 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 ...

  4. 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 ...

  5. 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 ...

  6. 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 ...

  7. Xamarin Error cannot find ‘aapt.exe’

    Problem:     solution:   A workaround is to copy your files to the old directory. Just copy the aapt ...

  8. 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 ...

  9. 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 ...

  10. 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 ...