[LeetCode] 647. Palindromic Substrings

最新推荐文章于 2022-05-31 21:15:53 发布

转载最新推荐文章于 2022-05-31 21:15:53 发布 · 58 阅读

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原文链接：https://segmentfault.com/a/1190000016419438

本文介绍了一种算法，用于计算给定字符串中回文子串的数量。通过两种方法实现：土办法使用两次循环检查所有可能的子串，中点延展法则从每个字符出发向两边扩展寻找回文串。

Problem

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.

Solution

土办法，两次循环 O(n^3)

class Solution {
    public int countSubstrings(String s) {
        int count = 0;
        for (int i = 0; i < s.length(); i++) {
            for (int j = i+1; j <= s.length(); j++) {
                String cur = s.substring(i, j);
                if (isPalindrome(cur)) count++;
            }
        }
        return count;
    }
    private boolean isPalindrome(String s) {
        if (s.length() == 1) return true;
        int i = 0, j = s.length()-1;
        while (i <= j) {
            if (s.charAt(i++) != s.charAt(j--)) return false;
        }
        return true;
    }
}

中点延展法


class Solution {
    int count = 0;
    public int countSubstrings(String s) {
        for (int i = 0; i < s.length(); i++) {
            extendFromIndex(s, i, i);
            extendFromIndex(s, i, i+1);
        }
        return count;
    }
    private void extendFromIndex(String s, int i, int j) {
        while (i >= 0 && j < s.length() && s.charAt(i--) == s.charAt(j++)) {
            count++;
        }
    }
}