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 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.
   
   
   

题目要求：给定一个string数组，遍历数组中的string字符串，如果该字符串的每一个字母都在键盘上的同一行，则将其保留，否则删除该string。

我的思路是用一个数组来记录每一个字母在哪一行，从0-25分别记录a/A-z/Z这26个字母的所在行数，所以声明数组如下：

int mark[26] = {2,3,3,2,1,2,2,2,1,2,2,2,3,3,1,1,1,1,2,1,1,3,1,3,1,3};

当我遍历每个字符串中的每个字符时，我利用ASCii码来确定该字母在数组中的索引，然后将前后两个字母比较，判断是否在同一行：

for (int j = 1; j < words[i].size(); j++) {                int index1 = (words[i][j-1] - 'a' >= 0) ? (words[i][j-1] - 'a') : (words[i][j-1] - 'A');                int index2 = (words[i][j] - 'a' >= 0) ? (words[i][j] - 'a') : (words[i][j] - 'A');                if (mark[index1] != mark[index2]) {                    isOneRow = false;                    break;                }            }

因为字符串中的字母有大小写之分，所以确定索引的时候要进行判断。

完整的代码如下：

class Solution {public:    vector<string> findWords(vector<string>& words) {        int mark[26] = {2,3,3,2,1,2,2,2,1,2,2,2,3,3,1,1,1,1,2,1,1,3,1,3,1,3};        vector<string> result;        for (int i = 0; i < words.size(); i++) {            bool isOneRow = true;            for (int j = 1; j < words[i].size(); j++) {                int index1 = (words[i][j-1] - 'a' >= 0) ? (words[i][j-1] - 'a') : (words[i][j-1] - 'A');                int index2 = (words[i][j] - 'a' >= 0) ? (words[i][j] - 'a') : (words[i][j] - 'A');                if (mark[index1] != mark[index2]) {                    isOneRow = false;                    break;                }            }            if (isOneRow)                result.push_back(words[i]);        }                return result;    }};