leetcode-面试题 16.18. 模式匹配

Posted 2020-12-12 oldby

 

 

 

 方法一：

class Solution:
    def patternMatching(self, pattern: str, value: str) -> bool:
        count_a = sum(1 for ch in pattern if ch == ‘a‘)
        count_b = len(pattern) - count_a
        if count_a < count_b:
            count_a, count_b = count_b, count_a
            pattern = ‘‘.join(‘a‘ if ch == ‘b‘ else ‘b‘ for ch in pattern)
        
        if not value:
            return count_b == 0
        if not pattern:
            return False
        
        for len_a in range(len(value) // count_a + 1):
            rest = len(value) - count_a * len_a
            if (count_b == 0 and rest == 0) or (count_b != 0 and rest % count_b == 0):
                len_b = 0 if count_b == 0 else rest // count_b
                pos, correct = 0, True
                value_a, value_b = None, None
                for ch in pattern:
                    if ch == ‘a‘:
                        sub = value[pos:pos+len_a]
                        if not value_a:
                            value_a = sub
                        elif value_a != sub:
                            correct = False
                            break
                        pos += len_a
                    else:
                        sub = value[pos:pos+len_b]
                        if not value_b:
                            value_b = sub
                        elif value_b != sub:
                            correct = False
                            break
                        pos += len_b
                if correct and value_a != value_b:
                    return True
        
        return False

方法二：正则

class Solution:
    def patternMatching(self, pattern: str, value: str) -> bool:
        # 边界情况处理
        if not pattern: return not value
        if len(pattern) == 1: return True

        # 构造正则表达式：重点是正则表达式的“后向引用”
        rega, regb = (‘\\1‘, ‘\\2‘) if pattern[0] == ‘a‘ else (‘\\2‘, ‘\\1‘)
        p = pattern.replace(‘a‘, ‘(\\w*)‘, 1).replace(‘b‘, ‘(\\w*)‘, 1).replace(‘a‘, rega).replace(‘b‘, regb)
        p = ‘^‘ + p + ‘$‘
        m = re.match(p, value)

        # 匹配到 && (模式长度为1 || 模式长度为2 && 两个模式不相同)
        return bool(m and (len(m.groups()) == 1 or m.groups()[0] != m.groups()[1]))

作者：dz-lee
链接：https://leetcode-cn.com/problems/pattern-matching-lcci/solution/7xing-dai-ma-zheng-ze-biao-da-shi-jie-fa-python3xi/
来源：力扣（LeetCode）
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