Leetcode 1147. Longest Chunked Palindrome Decomposition

最新推荐文章于 2023-04-13 21:16:02 发布

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最新推荐文章于 2023-04-13 21:16:02 发布

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本文介绍了一种基于贪婪算法的字符串匹配与分解方法，旨在找到一个字符串的最大可能分解数量，使得分解后的每个部分都能与其逆序部分相等。通过实例演示了如何实现这一算法，并详细解释了其工作原理。

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Return the largest possible k such that there exists a_1, a_2, ..., a_k such that:

Each a_i is a non-empty string;
Their concatenation a_1 + a_2 + ... + a_k is equal to text;
For all 1 <= i <= k, a_i = a_{k+1 - i}.

Example 1:

Input: text = "ghiabcdefhelloadamhelloabcdefghi"
Output: 7
Explanation: We can split the string on "(ghi)(abcdef)(hello)(adam)(hello)(abcdef)(ghi)".

Example 2:

Input: text = "merchant"
Output: 1
Explanation: We can split the string on "(merchant)".

Example 3:

Input: text = "antaprezatepzapreanta"
Output: 11
Explanation: We can split the string on "(a)(nt)(a)(pre)(za)(tpe)(za)(pre)(a)(nt)(a)".

Example 4:

Input: text = "aaa"
Output: 3
Explanation: We can split the string on "(a)(a)(a)".

Constraints:

text consists only of lowercase English characters.
1 <= text.length <= 1000

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DP is the first intuition. Approve the greedy algorithm:

Assume a longer match l_header = l_tail, a shorter match s_header = s_tail.Draw a map, this indicates l_header consists of (s_header, middle, s_header). So greedy algothm is enough:

class Solution:
    def longestDecomposition(self, text):
        l = len(text)
        res = 0
        
        r_text = text[::-1]
        cur,r_cur = '',''
        for i in range(0,l):
            cur = cur+text[i]
            r_cur = r_text[i]+r_cur
            if (cur == r_cur):
                res+=1
                cur,r_cur = '',''
        return res