zoukankan      html  css  js  c++  java
  • 500. Keyboard Row

    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:

    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.

    永远渴望,大智若愚(stay hungry, stay foolish)
  • 相关阅读:
    装饰着模式
    观察者模式
    策略模式
    nginx配置图片防盗链
    nginx配置文件详解( 看着好长,其实不长,看了就知道了,精心整理,有些配置也是没用到呢 )
    php引用计数的基本知识
    PHP运行模式
    CURL常用命令--update20151015
    memcache相同主域名下的session共享
    memcached命令行操作详解,命令选项的详细解释
  • 原文地址:https://www.cnblogs.com/h-hkai/p/9981358.html
Copyright © 2011-2022 走看看