Palindromic Substrings (M)

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:

1. The input string length won‘t exceed 1000.

题意

统计给定字符串中回文子串的个数。

思路

当两个回文子串的中心不同时，这两个子串必然是不同的。因此只需要遍历所有的中心位置(单个字符或者是字符之间的空隙)，累加不同中心位置对应的回文子串的个数即可得到答案。复杂度为\(O(N^2)\)。

也可使用动态规划进行判断：dp[i][j]表示子串i-j是否为回文串，则有如下递推式：

也可以使用马拉车算法 - Manacher‘s Algorithm，复杂度为\(O(N)\)。

代码实现

Java

中心判断

class Solution {
    public int countSubstrings(String s) {
        int count = 0;
        for (int i = 0; i < s.length(); i++) {
            count += countPalindrome(s, i, i) + countPalindrome(s, i, i + 1);
        }
        return count;
    }

    // 当p、q相等时，说明是以某个字符为中心扩展
    // 当p、q不等时，说明是以某个空隙为中心扩展
    private int countPalindrome(String s, int p, int q) {
        int count = 0;
        while (p >= 0 && q < s.length() && s.charAt(p--) == s.charAt(q++)) {
            count++;
        }
        return count;
    }
}

马拉车算法

class Solution {
    public int countSubstrings(String s) {
        if (s.isEmpty() || s == null) {
            return 0;
        }

        int count = 0;
        String t = transform(s);
        int[] radius = new int[t.length()];
        int center = 0;

        for (int i = 0; i < t.length(); i++) {
            int rightBound = center + radius[center];
            radius[i] = rightBound >= i ? Math.min(radius[2 * center - i], rightBound - i) : 0;
            while (i + radius[i] + 1 < t.length() && i - radius[i] - 1 >= 0
                    && t.charAt(i + radius[i] + 1) == t.charAt(i - radius[i] - 1)) {
                radius[i]++;
            }
            center = i + radius[i] > rightBound ? i : center;
            count += (radius[i] + 1) / 2;
        }
        return count;
    }

    private String transform(String s) {
        StringBuilder sb = new StringBuilder();
        for (char c : s.toCharArray()) {
            sb.append(‘#‘);
            sb.append(c);
        }
        sb.append(‘#‘);
        return sb.toString();
    }
}

JavaScript

/**
 * @param {string} s
 * @return {number}
 */
var countSubstrings = function (s) {
  let ans = 0
  const dp = new Array(s.length).fill(0).map(item => [])

  for (let len = 1; len <= s.length; len++) {
    for (let start = 0; start + len - 1 < s.length; start++) {
      const end = start + len - 1

      if (len === 1) {
        dp[start][end] = true
      } else if (len === 2) {
        dp[start][end] = s[start] === s[end]
      } else {
        dp[start][end] = s[start] === s[end] && dp[start + 1][end - 1]
      }

      if (dp[start][end]) ans++
    }
  }

  return ans
}

0647. Palindromic Substrings (M)

原文：https://www.cnblogs.com/mapoos/p/14586192.html