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。

示例 1：

输入： pattern = "abba", value = "dogcatcatdog"
输出： true

示例 2：

输入： pattern = "abba", value = "dogcatcatfish"
输出： false

示例 3：

输入： pattern = "aaaa", value = "dogcatcatdog"
输出： false

示例 4：

输入： pattern = "abba", value = "dogdogdogdog"
输出： true
解释： "a"="dogdog",b=""，反之也符合规则

提示：
0 <= len(pattern) <= 1000
0 <= len(value) <= 1000
你可以假设pattern只包含字母"a"和"b"，value仅包含小写字母。

• 先判断为空的极端条件；
• 然后开始记录a和b的数量，以及第一个a和b的坐标，根据坐标可以很容易找到两者不同长度时对应的字符串；
• 然后优先判断a和b的数量为1的情况；
• 最后开始整活，由二元一次方程可以构造a的长度和b长度的关系，所有只要遍历a的长度情况就行了。

class Solution {public boolean patternMatching(String pattern, String value) {int count_a = 0, count_b = 0;for (char ch: pattern.toCharArray()) {if (ch == 'a') {++count_a;} else {++count_b;}}if (count_a < count_b) {int temp = count_a;count_a = count_b;count_b = temp;char[] array = pattern.toCharArray();for (int i = 0; i < array.length; i++) {array[i] = array[i] == 'a' ? 'b' : 'a';}pattern = new String(array);}if (value.length() == 0) {return count_b == 0;}if (pattern.length() == 0) {return false;}for (int len_a = 0; count_a * len_a <= value.length(); ++len_a) {int rest = value.length() - count_a * len_a;if ((count_b == 0 && rest == 0) || (count_b != 0 && rest % count_b == 0)) {int len_b = (count_b == 0 ? 0 : rest / count_b);int pos = 0;boolean correct = true;String value_a = "", value_b = "";for (char ch: pattern.toCharArray()) {if (ch == 'a') {String sub = value.substring(pos, pos + len_a);if (value_a.length() == 0) {value_a = sub;} else if (!value_a.equals(sub)) {correct = false;break;}pos += len_a;} else {String sub = value.substring(pos, pos + len_b);if (value_b.length() == 0) {value_b = sub;} else if (!value_b.equals(sub)) {correct = false;break;}pos += len_b;}}if (correct && !value_a.equals(value_b)) {return true;}}}return false;}
}

class Solution {public boolean patternMatching(String pattern, String value) {int a = 0, b = 0;    //统计a与b的个数for(int i = 0; i < pattern.length(); ++i){if(pattern.charAt(i) == 'a') a++;else b++;}if(value.equals("")){//空串直接返回if(a*b == 0) return true;return false;//a与b对应的串不能同时为空}if(pattern.equals("")) return false;if(b == pattern.length()) a = b;    //只有a或只有b，统一处理if(a == pattern.length()){if(value.length()%a != 0) return false;int c = value.length()/a;String s = value.substring(0, c);//平均按照长度c断开匹配for(int i = 0; i < a; ++i){if(!s.equals(value.substring(c*i, c*(i + 1)))) return false;}return true;}List<List<Integer>> r = new ArrayList<>();//记录所有整数解的a与b的串的长度for(int i = 0; i <= value.length(); ++i){int temp = value.length() - a*i;if(temp < 0) break;if(temp%b == 0){List<Integer> res = new ArrayList<>();res.add(i);res.add(temp/b);r.add(res);}}int i = pattern.indexOf('a'), j = pattern.indexOf('b');for(List<Integer> v : r){int m = v.get(0), n = v.get(1), sum = 0;    //用sum累计下标值String s1 = value.substring(i*n, i*n + m), s2 = value.substring(j*m, j*m + n);if(s1.equals(s2)) continue;    //a与b对应的串不能相等boolean flag = true;for(int k = 0; k < pattern.length(); ++k){String s = pattern.charAt(k) == 'a'? s1: s2;int p = pattern.charAt(k) == 'a'? m: n;if(!s.equals(value.substring(sum, sum + p))){flag = false;break;}sum += p;}if(flag) return true;   //只要存在一组解，就为真}return false;}
}