[LeetCode] Keyboard Row 键盘行

Posted 2020-08-31 Grandyang

 

Given a List of words, return the words that can be typed using letters of alphabet on only one row\'s of American keyboard like the image below.

 

 

Example 1:

Input: ["Hello", "Alaska", "Dad", "Peace"]
Output: ["Alaska", "Dad"]

 

Note:

1. You may use one character in the keyboard more than once.
2. You may assume the input string will only contain letters of alphabet.

 

这道题给了我们一些单词，问哪些单词可以由键盘上的一行中的键符打出来，难度其实并不大。首先我们把键盘的三行字符分别保存到三个set中，然后遍历每个单词中的每个字符，分别看当前字符是否在三个集合中，如果在，对应的标识变量变为1，我们统计三个标识变量之和就知道有几个集合参与其中了，参见代码如下：

 

class Solution {
public:
    vector<string> findWords(vector<string>& words) {
        vector<string> res;
        unordered_set<char> row1{\'q\',\'w\',\'e\',\'r\',\'t\',\'y\',\'u\',\'i\',\'o\',\'p\'};
        unordered_set<char> row2{\'a\',\'s\',\'d\',\'f\',\'g\',\'h\',\'j\',\'k\',\'l\'};
        unordered_set<char> row3{\'z\',\'x\',\'c\',\'v\',\'b\',\'n\',\'m\'};
        for (string word : words) {
            int one = 0, two = 0, three = 0;
            for (char c : word) {
                if (c < \'a\') c += 32;
                if (row1.count(c)) one = 1;
                if (row2.count(c)) two = 1;
                if (row3.count(c)) three = 1;
                if (one + two + three > 1) break;
            }
            if (one + two + three == 1) res.push_back(word);
        }
        return res;
    }
};

 

参考资料：

https://discuss.leetcode.com/topic/77754/java-1-line-solution-via-regex-and-stream

 

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