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N647_PalindromicSubstrings.java
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N647_PalindromicSubstrings.java
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/**
* @Author: parallelline1996
* @Email: chenyu1996a@163.com
* @Date: 2019/12/18 23:00
*/
package page7;
/**
* https://leetcode.com/problems/palindromic-substrings/
*
* 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.
*/
public class N647_PalindromicSubstrings {
public static void main(String[] args) {
System.out.println(new N647_PalindromicSubstrings().countSubstrings("abbbba"));
System.out.println(new N647_PalindromicSubstrings().countSubstrings("aaa"));
System.out.println(new N647_PalindromicSubstrings().countSubstrings("abc"));
System.out.println(new N647_PalindromicSubstrings().countSubstrings("a"));
System.out.println(new N647_PalindromicSubstrings().countSubstrings1("abbbba"));
System.out.println(new N647_PalindromicSubstrings().countSubstrings1("aaa"));
System.out.println(new N647_PalindromicSubstrings().countSubstrings1("abc"));
System.out.println(new N647_PalindromicSubstrings().countSubstrings1("a"));
}
public int countSubstrings(String s) {
int num = 0;
char[] a = s.toCharArray();
// 注意,这里需根据子串长度的奇偶分两种情况
for (int i = 0; i < s.length(); i++) {
num += count(a, i, i);
num += count(a, i, i + 1);
}
return num;
}
private int count(char[] a, int left, int right) {
int num = 0;
// 当某个子串是回文子串时,向左右两侧扩展,直至不再是回文子串或超过数据越界
while (left >= 0 && right < a.length && a[left] == a[right]) {
num++;
left--;
right++;
}
return num;
}
public int countSubstrings1(String s) {
if (s == null) {
return 0;
}
int num = 0;
// dp[i][j]为true代表:以下标i为起点,下标j为终点的子串为回文子串
boolean dp[][] = new boolean[s.length()][s.length()];
for (int i = s.length() - 1; i >= 0; i--) {
for (int j = i; j < s.length(); j++) {
if (s.charAt(i) == s.charAt(j) && ((j - i <= 2) || (dp[i + 1][j - 1]))) {
dp[i][j] = true;
num++;
}
}
}
return num;
}
//dp[i,j] means whether substring (i,j) of s is a palindrome
}