Problem

Given a string, your task is to count how many palindromic substrings in this string.

The substrings with different start indexes or end indexes are counted as different substrings even they consist of same characters.

Example 1:

Input: "abc"
Output: 3
Explanation: Three palindromic strings: "a", "b", "c".
 

Example 2:

Input: "aaa"
Output: 6
Explanation: Six palindromic strings: "a", "a", "a", "aa", "aa", "aaa".

Note:The input string length won't exceed 1000.

Solution

土辦法，兩次迴圈 O(n^3)

class Solution {
    public int countSubstrings(String s) {
        int count = 0;
        for (int i = 0; i < s.length(); i++) {
            for (int j = i+1; j <= s.length(); j++) {
                String cur = s.substring(i, j);
                if (isPalindrome(cur)) count++;
            }
        }
        return count;
    }
    private boolean isPalindrome(String s) {
        if (s.length() == 1) return true;
        int i = 0, j = s.length()-1;
        while (i <= j) {
            if (s.charAt(i++) != s.charAt(j--)) return false;
        }
        return true;
    }
}
 

中點延展法

class Solution {
    int count = 0;
    public int countSubstrings(String s) {
        for (int i = 0; i < s.length(); i++) {
            extendFromIndex(s, i, i);
            extendFromIndex(s, i, i+1);
        }
        return count;
    }
    private void extendFromIndex(String s, int i, int j) {
        while (i >= 0 && j < s.length() && s.charAt(i--) == s.charAt(j++)) {
            count++;
        }
    }
}