LeetCode 647. Palindromic Substrings
原题链接在这里:https://leetcode.com/problems/palindromic-substrings/description/
题目:
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.
题解:
对于String s的每一位向左右延展看是不是palindrome. 延展时有奇数偶数个char的substring两种情况.
Time Complexity: O(n^2). 延展用时2*(1+2+ ... + n/2 + ... +2+1) = O(n^2).
Space: O(1).
AC Java:
1 class Solution { 2 int count = 0; 3 4 public int countSubstrings(String s) { 5 if(s == null || s.length() == 0){ 6 return 1; 7 } 8 9 for(int i = 0; i<s.length(); i++){ 10 // 有奇数char的substring 11 extendSubstring(s, i, i); 12 // 有偶数char的substring 13 extendSubstring(s, i, i+1); 14 } 15 16 return count; 17 } 18 19 private void extendSubstring(String s, int l, int r){ 20 while(l>=0 && r<s.length() && s.charAt(l)==s.charAt(r)){ 21 count++; 22 l--; 23 r++; 24 } 25 } 26 }