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647. 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.
参考资料
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.
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/**
* Runtime: 1 ms, faster than 100.00% of Java online submissions for Palindromic Substrings.
* Memory Usage: 37.8 MB, less than 11.39% of Java online submissions for Palindromic Substrings.
*
* Copy from: https://leetcode-cn.com/problems/palindromic-substrings/solution/hui-wen-zi-chuan-by-leetcode/[回文子串 - 回文子串 - 力扣（LeetCode）]
*
* @author D瓜哥 · https://www.diguage.com
* @since 2020-01-31 23:17
*/
public int countSubstrings(String s) {
char[] A = new char[2 * s.length() + 3];
A[0] = '@';
A[1] = '#';
A[A.length - 1] = '$';
int j = 2;
for (char c : s.toCharArray()) {
A[j++] = c;
A[j++] = '#';
}
int[] Z = new int[A.length];
int center = 0, right = 0;
for (int i = 1; i < Z.length - 1; i++) {
if (i < right) {
Z[i] = Math.min(right - i, Z[2 * center - i]);
}
while (A[i + Z[i] + 1] == A[i - Z[i] - 1]) {
Z[i]++;
}
if (i + Z[i] > right) {
center = i;
right = i + Z[i];
}
}
int result = 0;
for (int i : Z) {
result += (i + 1) / 2;
}
return result;
}