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500. Keyboard Row

时间:2018-11-19 11:01:47      阅读:173      评论:0      收藏:0      [点我收藏+]

标签:分享   false   string   rac   dad   app   other   order   can   

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.

 

技术分享图片

 

Example:

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.

 

Approach #1: C++.

class Solution {
public:
    vector<string> findWords(vector<string>& words) {
        vector<string> ans;
        vector<unordered_set<char>> temp = {
                                                {‘q‘, ‘w‘, ‘e‘, ‘r‘, ‘t‘, ‘y‘,‘u‘, ‘i‘, ‘o‘, ‘p‘},
                                                {‘a‘, ‘s‘, ‘d‘, ‘f‘, ‘g‘, ‘h‘, ‘j‘, ‘k‘, ‘l‘},
                                                {‘z‘, ‘x‘, ‘c‘, ‘v‘, ‘b‘, ‘n‘, ‘m‘}
                                           };
        int size = words.size();
        
        for (int i = 0; i < size; ++i) {
            int flag1 = -1;
            bool ant = false;
            int len = words[i].length();
            for (int j = 0; j < len; ++j) {
                int flag2 = flag1;
                if (words[i][j] > ‘z‘) words[i][j] -= 65;
                if (temp[0].count(words[i][j])) flag1 = 0;
                if (temp[1].count(words[i][j])) flag1 = 1;
                if (temp[2].count(words[i][j])) flag1 = 2;
                if (flag2 >= 0 && flag1 != flag2) {
                    ant = true;
                    break;
                }
            }
            if (!ant) ans.push_back(words[i]);
        }
        
        return ans;
    }
};

  

Approach #2: Java.

class Solution {
    public String[] findWords(String[] words) {
        String[] strs = {"QWERTYUIOP", "ASDFGHJKL", "ZXCVBNM"};
        Map<Character, Integer> map = new HashMap<>();
        for (int i = 0; i < strs.length; ++i) {
            for (char c: strs[i].toCharArray()) {
                map.put(c, i);
            }
        }
        List<String> res = new LinkedList<>();
        for (String w : words) {
            if (w.equals("")) continue;
            int index = map.get(w.toUpperCase().charAt(0));
            for (char c : w.toUpperCase().toCharArray()) {
                if (map.get(c) != index) {
                    index = -1;
                    break;
                }
            }
            if (index != -1) res.add(w);
        }
        return res.toArray(new String[0]);
    }
}

  

Approach #3: Python.

class Solution(object):
    def findWords(self, words):
        """
        :type words: List[str]
        :rtype: List[str]
        """
        line1, line2, line3 = set("qwertyuiop"), set("asdfghjkl"), set("zxcvbnm")
        ret = []
        for word in words:
            w = set(word.lower())
            if w.issubset(line1) or w.issubset(line2) or w.issubset(line3):
                ret.append(word)
        return ret

  

Analysis:

In the approach one, I use a vector<unordered_set<char>> to contion the keyboard row. Then checking the word‘s characters is only be contioned in one keyboard row. I use tow flags with flag1 and flag2 to mark the previous character and the current character, if they are same with each other always, we can push it to the return vector.

 

500. Keyboard Row

标签:分享   false   string   rac   dad   app   other   order   can   

原文地址:https://www.cnblogs.com/ruruozhenhao/p/9981358.html

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