647.Palindromic-Substrings.md

647. Palindromic Substrings

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题目地址

https://leetcode.com/problems/palindromic-substrings/

题目描述

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.

代码

Approach #1

class Solution {
  int count = 0;
  public int countSubstrings(String s) {
    if (s == null || s.length() == 0)		return 0;
    
    for (int i = 0; i < s.length(); i++) {
      extendPalindrome(s, i, i);
      extendPalindrome(s, i, i + 1);
    }
    
    return count;
  }
  
  private void extendPalindrome(String s, int left, int right) {
    while (left >= 0 && right < s.length() 
           && s.charAt(left) == s.charAt(right)) {
      count++;
      left--;
      right++;
    }
  }
}

Approach #1 Expand Around Center

Time O(N^2) Space O(1)

class Solution {
  public int countSubstrings(String S) {
		int N = S.length(), ans = 0;
    for (int center = 0; center <= 2 * N - 1; center++) {
      int left = center / 2;
      int right = left + center % 2;
      while (left >= 0 && right < N 
             && S.charAt(left) == S.charAt(right)) {
        
        ans++;
        left--;
        right++;
      }
    }
    
    return ans;
  }
}

Approach #2 Manacher's Algorithm

Time O(N) Space O(N)

class Solution {
  public int countSubstrings(String S) {
    char[] A = new char[2 * S.length() + 3];
    A[0] = '@';
    A[1] = '#';
    A[A.length - 1] = '$';
    int t = 2;
    for (char c: S.toCharArray()) {
      A[t++] = c;
      A[t++] = '#';
    }
    
    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 ans = 0;
    for (int v: Z) {
      ans += (v + 1) / 2;
    }
    
    return ans;
  }
}