Description
A query word matches a given pattern if we can insert lowercase letters to the pattern word so that it equals the query. (We may insert each character at any position, and may insert 0 characters.)
Given a list of queries, and a pattern, return an answer list of booleans, where answer[i] is true if and only if queries[i] matches the pattern.
Example 1:
Input: queries = ["FooBar","FooBarTest","FootBall","FrameBuffer","ForceFeedBack"], pattern = "FB" Output: [true,false,true,true,false] Explanation: "FooBar" can be generated like this "F" + "oo" + "B" + "ar". "FootBall" can be generated like this "F" + "oot" + "B" + "all". "FrameBuffer" can be generated like this "F" + "rame" + "B" + "uffer". Example 2:
Input: queries = ["FooBar","FooBarTest","FootBall","FrameBuffer","ForceFeedBack"], pattern = "FoBa" Output: [true,false,true,false,false] Explanation: "FooBar" can be generated like this "Fo" + "o" + "Ba" + "r". "FootBall" can be generated like this "Fo" + "ot" + "Ba" + "ll". Example 3:
Input: queries = ["FooBar","FooBarTest","FootBall","FrameBuffer","ForceFeedBack"], pattern = "FoBaT" Output: [false,true,false,false,false] Explanation: "FooBarTest" can be generated like this "Fo" + "o" + "Ba" + "r" + "T" + "est". Note:
1 <= queries.length <= 1001 <= queries[i].length <= 1001 <= pattern.length <= 100- All strings consists only of lower and upper case English letters.
这道题说是若要一个 query 单词可以匹配一个模式 pattern 串的话，需要在在 pattern 中加入若干小写字母使其和 query 单词完全相同，现在给了一个单词数组和一个 pattern 串，问数组中的每个 query 单词是否都可以成功匹配。根据题目中的给的例子可以分析出，pattern 串中是可以有小写字母的，但主要是要其中的每个大写字母能匹配上，所以核心是要匹配对应位置的大写字母。由于 query 单词之间没有什么联系，所以每个 query 单词和 pattern 串的匹配方式都是相同的。用一个变量i来指向当前检测到 pattern 串中的位置，用个布尔变量 valid 来标记当前 query 是否能匹配。遍历 query 单词中的字符，若 i 小于 pattern 的长度，且当前字符等于 pattern[i]，则i自增1；否则若当前字符是个大写字符，则说明无法匹配，标记 valid 为 false，并 break 掉 for 循环。最后若 valid 为 true，且i正好等于n时，加入 true 到结果 res，否则加入 false，参见代码如下：
解法一：
class Solution { public: vector<bool> camelMatch(vector<string>& queries, string pattern) { vector<bool> res; for (string query : queries) { int i = 0, n = pattern.size(); bool valid = true; for (char c : query) { if (i < n && c == pattern[i]) { ++i; } else if (c >= 'A' && c <= 'Z') { valid = false; break; } } res.push_back(valid && i == n); } return res; } }; 我们也可以将验证部分放入一个子函数中，整体结构更加清晰一些，但整体思路都是一样的，参见代码如下：
解法二：
class Solution { public: vector<bool> camelMatch(vector<string>& queries, string pattern) { vector<bool> res; for (string query : queries) { res.push_back(helper(query, pattern)); } return res; } bool helper(string query, string pattern) { int i = 0, n = pattern.size(); for (char c : query) { if (i < n && c == pattern[i]) { ++i; } else if (c >= 'A' && c <= 'Z') { return false; } } return i == n; } }; Github 同步地址:
参考资料：
https://leetcode.com/problems/camelcase-matching/
https://leetcode.com/problems/camelcase-matching/discuss/270006/Java-Easy-Two-Pointers