【问题描述】[中等]

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。

【解答思路】

枚举

时间复杂度

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;}
}

【总结】

1. 有思路要实现 注意细节 Talk is cheap，show me the code.

2.双字符串匹配问题

2.1 边界问题 逐个是否为空
2.2 两个之间的长度 细节比较
2.3 匹配问题 可以转化为出现次数*长度

3.细节

字符串转字符数组 & 处理 & 字符数组转字符串

char[] array = pattern.toCharArray();for (int i = 0; i < array.length; i++) {array[i] = array[i] == 'a' ? 'b' : 'a';}pattern = new String(array);

转载链接：https://leetcode-cn.com/problems/pattern-matching-lcci/solution/mo-shi-pi-pei-by-leetcode-solution/