1352. Product of the Last K Numbers

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. 1 <= queries.length <= 100
2. 1 <= queries[i].length <= 100
3. 1 <= pattern.length <= 100
4. 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

https://leetcode.com/problems/camelcase-matching/discuss/270029/JavaC%2B%2BPython-Check-Subsequence-and-Regax

LeetCode All in One 题目讲解汇总(持续更新中…)

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