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

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