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)。
1 #include<bits/stdc++.h> 2 using namespace std; 3 const int MAX_N=1e5+10; 4 char s[MAX_N]; 5 int len; 6 int opp(int j, int x){ 7 if(x==0){ 8 return len+1-j; 9 } 10 if(j<x){ 11 return len-j<x? len-j: len-j+1; 12 } 13 else{ 14 return len+2-j; 15 } 16 } 17 bool ok(int x){ 18 int tmp=x?(len-1)>>1:len>>1; 19 for(int i=0, j=1; i<tmp; i++, j++){ 20 if(j==x){ 21 j++; 22 } 23 if(s[j]!=s[opp(j, x)]){ 24 return false; 25 } 26 } 27 return true; 28 } 29 int main(){ 30 int T; 31 scanf("%d", &T); 32 while(T--){ 33 scanf("%s", s+1); 34 len=strlen(s+1); 35 for(int i=0; i<=len; i++){ 36 if(ok(i)){ 37 printf("%d ", i-1); 38 break; 39 } 40 } 41 } 42 return 0; 43 }
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