[LeetCode] 647. Palindromic Substrings 回文子字符串

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.

给了一个字符串,计算有多少个回文子字符串,不同index的都算作不同的子字符串。

解法1: DP

解法2: Manacher's Algorithm

Python: DP

class Solution(object):
    def countSubstrings(self, s):
        """
        :type s: str
        :rtype: int
        """
        n = len(s)
        count = 0
        start, end, maxL = 0, 0, 0
        dp = [[0] * n for _ in range(n)]
        for i in range(n):
            for j in range(i):
                dp[j][i] = (s[j] == s[i]) & ((i - j < 2) | dp[j + 1][i - 1])
                if dp[j][i]:
                    count += 1
            dp[i][i] = 1
            count += 1
        return count

Python: Manacher's Algorithm

class Solution(object):
    def countSubstrings(self, s):
        """
        :type s: str
        :rtype: int
        """
        def manacher(s):
            s = '^#' + '#'.join(s) + '#$'
            P = [0] * len(s)
            C, R = 0, 0
            for i in xrange(1, len(s) - 1):
                i_mirror = 2*C-i
                if R > i:
                    P[i] = min(R-i, P[i_mirror])
                while s[i+1+P[i]] == s[i-1-P[i]]:
                    P[i] += 1
                if i+P[i] > R:
                    C, R = i, i+P[i]
            return P
        return sum((max_len+1)/2 for max_len in manacher(s))

C++:

class Solution {
public:
    int countSubstrings(string s) {
        if (s.empty()) return 0;
        int n = s.size(), res = 0;
        for (int i = 0; i < n; ++i) {
            helper(s, i, i, res);
            helper(s, i, i + 1, res);
        }
        return res;
    }
    void helper(string s, int i, int j, int& res) {
        while (i >= 0 && j < s.size() && s[i] == s[j]) {
            --i; ++j; ++res;
        }
    }
};

C++:

class Solution {
public:
    int countSubstrings(string s) {
        int n = s.size(), res = 0;
        vector<vector<bool>> dp(n, vector<bool>(n, false));
        for (int i = n - 1; i >= 0; --i) {
            for (int j = i; j < n; ++j) {
                dp[i][j] = (s[i] == s[j]) && (j - i <= 2 || dp[i + 1][j - 1]);
                if (dp[i][j]) ++res;
            }
        }
        return res;
    }
};

  

类似题目:

[LeetCode] 5. Longest Palindromic Substring 最长回文子串

[LeetCode] 9. Palindrome Number 验证回文数字

[LeetCode] 125. Valid Palindrome 有效回文

[LeetCode] 516. Longest Palindromic Subsequence 最长回文子序列

  

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posted @ 2018-08-25 08:11  轻风舞动  阅读(498)  评论(0编辑  收藏  举报