【leetcode】1147. Longest Chunked Palindrome Decomposition

题目如下：

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_kis 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

解题思路：本题不算太难，我的方法是贪心算法+双指针。首先引入head和tail两个变量，分别等于text[0]和text[-1]。如果head等于tail，表示这两者可以组成回文段的两部分，再令head等于text[1]，tail等于text[-2]；如果两者不相等，令head = head + text[0]，tail = text[-2] + tail，直到head 等于tail为止。原则就是每遇到head等于tail的情况，表示这两段是回文段的一部分，重置head 和tail的值。

代码如下：

class Solution(object):
    def longestDecomposition(self, text):
        """
        :type text: str
        :rtype: int
        """
        res = 0
        head_inx = 0
        tail_inx = len(text) - 1
        head = ''
        tail = ''
        while head_inx <= tail_inx and head_inx < len(text) and tail_inx >= 0:
            if head == '' and tail == '':
                head = text[head_inx]
                tail = text[tail_inx]
                head_inx += 1
                tail_inx -= 1
            elif head == tail:
                res += 2
                head = text[head_inx]
                tail = text[tail_inx]
                head_inx += 1
                tail_inx -= 1
            else:
                #head_inx += 1
                #tail_inx -= 1
                head = head + text[head_inx]
                tail = text[tail_inx] + tail
                head_inx += 1
                tail_inx -= 1
        res += 2 if  head == tail and head_inx - len(head) != tail_inx + len(tail) else 1
        return res if res != 0 else 1