Week4 - 500.Keyboard Row & 557.Reverse Words in a String III

500.Keyboard Row & 557.Reverse Words in a String III

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.

#include<string>
#include<iostream>
#include<vector>
using namespace std;
static int alphabet[] = { 2,1,1,2,3,2,2,2,3,2,2,2,1,1,3,3,3,3,2,3,3,1,3,1,3,1,0,0,0,0,0,0,2,1,1,2,3,2,2,2,3,2,2,2,1,1,3,3,3,3,2,3,3,1,3,1,3,1 };
class Solution {
public:
  vector<string> findWords(vector<string>& words) {
    vector<string> result;
    for (int i = 0; i < words.size(); i++) {
      int flag = 0;
      bool same = false;
      if (words[i].size() > 0) flag = alphabet[(int)words[i][0] - 'A'];
      same = true;
      for (int j = 0; j < words.at(i).length(); j++) {
        if (alphabet[words[i][j] - 'A'] != flag) {
          same = false;
          break;
        }
      }
      if (same == true) result.push_back(words[i]);
    }
    return result;
  }
};

557.Reverse Words in a String III
Given a string, you need to reverse the order of characters in each word within a sentence while still preserving whitespace and initial word order.
Example 1:
Input: "Let's take LeetCode contest"
Output: "s'teL ekat edoCteeL tsetnoc"
Note: In the string, each word is separated by single space and there will not be any extra space in the string.

#include<string>
#include<iostream>
#include<vector>
#include<list>
using namespace std;
class Solution {
public:
  string reverseWords(string s) {
    string result;
    int flag = 0;
    for (int i = 0; i < s.length(); i++) {
      if (s[i] == ' ') {
        for (int j = i - 1; j >= flag; j--) {
          result.push_back(s[j]);
        }
        result.push_back(' ');
        flag = i + 1;
      }
    }
    for (int j = s.length()-1; j >= flag; j--) {
      result.push_back(s[j]);
    }
    return result;
  }
};