标签：vat task start value art count new tput for while
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".
Solution 1
class Solution { public int countSubstrings(String s) { int len = s.length(); int res = 0; boolean [][] isPalin = new boolean[len][len]; for (int i = 0; i < len; i++) { for (int j = 0; j <= i; j++) { if (s.charAt(i) == s.charAt(j) && (i <= j + 2 || isPalin[i - 1][j + 1])) { isPalin[i][j] = true; res += 1; } } } return res; } }
Solution 2
class Solution { public int countSubstrings(String s) { int len = s.length(); int[] count = new int[1]; for (int i = 0; i < len; i++) { getPalin(s, i, i, count); getPalin(s, i, i + 1, count); } return count[0]; } private void getPalin(String s, int left, int right, int[] count) { // need to ensure left, right value for while while (left >= 0 && right < s.length() && s.charAt(left) == s.charAt(right)) { count[0] += 1; left -= 1; right += 1; } } }
[LC] 647. Palindromic Substrings
标签：vat task start value art count new tput for while
原文地址：https://www.cnblogs.com/xuanlu/p/12591012.html
