傳送門：?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.

讀題時需注意：

題目中先說 “There will always be a valid solution. ”，然后才說“In case the string is already a palindrome, then?-1?is also a valid answer along with possible indices.”。注意體會這句話，我們首先應注意到，即使輸入的字符串S是個回文串，也可以刪除某個字母使其仍為回文串。如果|S|為奇數，則刪除中間那個字母，結果串仍為回文串。如果|S|為偶數則刪除中間兩個相等字符中的任一個，結果串也回文。

完全暴力的解法：

枚舉要刪除的字母，檢查結果串是否回文。復雜度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\n", i-1); 38 break; 39 } 40 } 41 } 42 return 0; 43 }

只是這解法過于暴力，TLE。

下面就要引入這道題給我的最大啟示了：

尋找有助于簡化問題的必要條件

考慮一下上面的單純暴力算法有那些冗余計算。

首先必須指出一個問題：優(yōu)化算法的途徑是充分考慮問題的特殊性。

其次要注意到：題目要求的是存在性判別，上面的算法枚舉被刪除字符的位置是無可厚非的。

接著考慮一下使上面的算法達到最壞情況的數據：

aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaab

在這種情況下，上述算法必須枚舉到最后一個字符才能確定答案。

我們不難發(fā)現(xiàn)一個問題

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轉載于:https://www.cnblogs.com/Patt/p/4467609.html

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