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 <= 100
1 <= queries[i].length <= 100
1 <= 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:
https://github.com/grandyang/leetcode/issues/1023
参考资料:
https://leetcode.com/problems/camelcase-matching/
https://leetcode.com/problems/camelcase-matching/discuss/270006/Java-Easy-Two-Pointers