Write a function:
class Solution { public int solution(String S); }
that, given a string S, returns the index (counting from 0) of a character such that the part of the string to the left of that character is a reversal of the part of the string to its right. The function should return −1 if no such index exists.
Note: reversing an empty string (i.e. a string whose length is zero) gives an empty string.
For example, given a string:
"racecar"
the function should return 3, because the substring to the left of the character "e" at index 3 is "rac", and the one to the right is "car".
Given a string:
"x"
the function should return 0, because both substrings are empty.
Write an efficient algorithm for the following assumptions:
- the length of string S is within the range [0..2,000,000].
// you can also use imports, for example:
// import java.util.*;
// you can use System.out.println for debugging purposes, e.g.
// System.out.println("this is a debug message");
class Solution {
public int solution(String S) {
// write your code in Java SE 8
int i = 0;
int j = S.length()-1;
while (i<j){
if (S.charAt(i)==S.charAt(j)){
i++;
j--;
}
else break;
}
if (i==j) return i;
else return -1;
}
}
// you can also use imports, for example:
// import java.util.*;
// you can use System.out.println for debugging purposes, e.g.
// System.out.println("this is a debug message");
class Solution {
public int solution(String S) {
// write your code in Java SE 8
int i = 0;
int j = S.length()-1;
while (i<j){
if (S.charAt(i)==S.charAt(j)){
i++;
j--;
}
else break;
}
if (i==j) return i;
else return -1;
}
}
// you can also use imports, for example:
// import java.util.*;
// you can use System.out.println for debugging purposes, e.g.
// System.out.println("this is a debug message");
class Solution {
public int solution(String S) {
// write your code in Java SE 8
int i = 0;
int j = S.length()-1;
while (i<j){
if (S.charAt(i)==S.charAt(j)){
i++;
j--;
}
else break;
}
if (i==j) return i;
else return -1;
}
}
The solution obtained perfect score.