LeetCode: 500 Keyboard Row (easy)

题目:

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.

代码:

class Solution {
public:
    vector<string> findWords(vector<string>& words) {
        map<char, int> m;
        vector<char> s1 = {'q', 'w', 'e', 'r', 't', 'y', 'u', 'i', 'o', 'p'};
        vector<char> s2 = {'a', 's', 'd', 'f', 'g', 'h', 'j', 'k', 'l'};
        vector<char> s3 = {'z', 'x', 'c', 'v', 'b', 'n', 'm'};
        for (auto c : s1)
            m.insert(pair<char, int> (c, 1));
        for (auto c : s2)
            m.insert(pair<char, int> (c, 2));
        for (auto c : s3)
            m.insert(pair<char, int> (c, 3));
                           
        vector<string> result;
        for (auto w : words){
            bool b = 1; #判断是否在一行
            char first = tolower(w[0]);
            int i = m[first];
            for (auto c : w ){
                c = tolower(c);
                int j = m[c];
                if (j != i){
                    b = 0;
                    break;
                }
            }
            if (b)
                result.push_back(w);
         }         
        return result;
    }
};

别人的:

 1 class Solution {
 2 public:
 3     vector<string> findWords(vector<string>& words) {
 4         
 5         vector<string> ans;
 6         unordered_set<char> row1{ 'q', 'w', 'e', 'r', 't', 'y', 'u', 'i', 'o', 'p' };
 7         unordered_set<char> row2{ 'a', 's', 'd', 'f', 'g', 'h', 'j', 'k', 'l' };
 8         unordered_set<char> row3{ 'z', 'x', 'c', 'v', 'b', 'n', 'm' };
 9         
10         for(string word:words){
11             int one = 0, two = 0, three = 0;
12             for(char c:word){
13                 if(row1.count(c)) one = 1;
14                 if(row2.count(c)) two = 1;
15                 if(row3.count(c)) three = 1;
16                 if(one+two+three > 1) break;
17             }
18             
19             if(one + two + three == 1) ans.push_back(word);
20         }
21         
22         return ans;
23          
24     }
25 };
 ASCII编码:http://www.doc88.com/p-951217962470.html

posted on 2017-08-21 15:19  玲珑子  阅读(153)  评论(0编辑  收藏  举报

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