模式匹配 A

题目

你有两个字符串，即pattern和value。 pattern字符串由字母"a"和"b"组成，用于描述字符串中的模式。例如，字符串"catcatgocatgo"匹配模式"aabab"（其中"cat"是"a"，“go"是"b”），该字符串也匹配像"a"、"ab"和"b"这样的模式。但需注意"a"和"b"不能同时表示相同的字符串。编写一个方法判断value字符串是否匹配pattern字符串。

解题思路

简单的思路：设$l_v$为value的长度，$l_a$为a对应模式的长度，$l_b$为b对应模式的长度，$coun{t}_a$为a对应模式的个数，$coun{t}_b$为b对应模式的个数。则有：
$$coun{t}_a\times l_a+coun{t}_b\times l_b=l_v$$
则在已知$coun{t}_a$时，以$l_a$和$l_b$为未知量得到二元一次方程，并且$l_a\in [0,\frac{l_v}{coun{t}_a}+1)$。
$l_a$处于这个区间以保证$l_b$大于等于0。接下来遍历$l_a$，得到分割后的子串是否满足要求。

0.统计a,b个数

统计pattern中a,b的个数

count_a = sum(1 for item in pattern if item == "a")
count_b = len(pattern)-count_a

1.特判

1. 若a的个数小于b的个数，调换a、b以保证a最少出现1次，继而枚举a

if count_a < count_b:
    count_a, count_b = count_b, count_a
    pattern = "".join("a" if item == "b" else "b" for item in pattern)

2. 若value为空,则要求pattern 中要么为空，要么只有a,等价于b的个数为0

if not value: return count_b == 0

3. 若pattern为空而value不为空，则无法匹配。

if not pattern and value: return False

2.对于pattern和value都不为空时，两重遍历

for len_a in range(len(value)//count_a + 1):#遍历a的可能长度
    leftForb = len(value) - count_a * len_a#可能的b的总长度
    #看当前的分法是否满足要求（len(b)为大于等于0的整数）
    if (count_b == 0 and leftForb == 0) or (count_b != 0 and leftForb % count_b == 0):
        len_b = 0 if count_b == 0 else leftForb // count_b
        idx, Judge = 0, True
        value_a, value_b = None, None
        for ch in pattern:#对于确定a,b长度进行遍历模式
            if ch == "a":#对模式a的判断
                subChr = value[idx:idx+len_a]
                if not value_a:
                    value_a = subChr
                elif value_a != subChr:
                    Judge = False
                    break
                idx += len_a
            else:#对模式b的判断
                subChr = value[idx:idx+len_b]
                if not value_b:
                    value_b = subChr
                elif value_b != subChr:
                    Judge = False
                    break
                idx += len_b
        if Judge and value_a != value_b:#找到满足的value_a,value_b且两者不相等，返回T
            return True

代码

class Solution:
    def patternMatching(self, pattern: str, value: str) -> bool:
        #统计a、b的个数
        count_a = sum(1 for item in pattern if item == "a")
        count_b = len(pattern)-count_a
        #若a的个数小于b的个数，调换a、b以保证a最少出现1次，继而枚举a
        if count_a < count_b:
            count_a, count_b = count_b, count_a
            pattern = "".join("a" if item == "b" else "b" for item in pattern)
        #若value为空,则要求pattern 中要么为空，要么只有a,等价于b的个数为0
        if not value: return count_b == 0
        #若pattern为空而value不为空，则无法匹配
        if not pattern and value: return False
        #对于pattern和value都不为空时，遍历
        for len_a in range(len(value)//count_a + 1):
            leftForb = len(value) - count_a * len_a#可能的b的总长度
            #看当前的分法是否满足要求（len(b)为大于等于0的整数）
            if (count_b == 0 and leftForb == 0) or (count_b != 0 and leftForb % count_b == 0):
                len_b = 0 if count_b == 0 else leftForb // count_b
                idx, Judge = 0, True
                value_a, value_b = None, None
                for ch in pattern:
                    if ch == "a":#对模式a的判断
                        subChr = value[idx:idx+len_a]
                        if not value_a:
                            value_a = subChr
                        elif value_a != subChr:
                            Judge = False
                            break
                        idx += len_a
                    else:#对模式b的判断
                        subChr = value[idx:idx+len_b]
                        if not value_b:
                            value_b = subChr
                        elif value_b != subChr:
                            Judge = False
                            break
                        idx += len_b
                if Judge and value_a != value_b:#找到满足的value_a,value_b且两者不相等，返回T
                    return True
        return False#遍历完都没找到，返回F