Title source

• 423. Reconstructing Chinese from English

Title Description

title

We know , Each number corresponds to an English word ,“zero”, “one”, “two”, “three”, “four”, “five”, “six”, “seven”, “eight”, “nine”, If you look at each English word in these words , We can find out ：

• The characters of some words are unique
  • such as z, Only in zero in
  • also w,u,x,g These four words , They only appear in two,four,six,eight in
  • Then the number of these five numbers can be determined
• Some words appear many times
  • Because of the o The words are zero,two,four,one, The first three of them have been identified , that one You will know the number of
  • Because of the h The words are eight,three, among eight The number is known , that three Just know the number of
  • Because of the f The words are four,five, among four The number is known , that five Just know the number of ;
  • Because of the s The words are six,seven, among six The number is known , that seven Just know the number of ;
  • Because of the i The words are six,eight,five,nine, The first three of them have been identified , that nine Just know the number of

Ideas ： First count the number of times each character appears , Then calculate the number of times each word appears , Then you can rebuild .

class Solution {
    
public:
    string originalDigits(string s) {
    
        string res = "";
        vector<string> words{
    "zero", "two", "four", "six", "eight", "one", "three", "five", "seven", "nine"};
        vector<int> nums{
    0, 2, 4, 6, 8, 1, 3, 5, 7, 9}, counts(26, 0);
        vector<char> chars{
    'z', 'w', 'u', 'x', 'g', 'o', 'h', 'f', 's', 'i'};
        for (char c : s) ++counts[c - 'a'];
        for (int i = 0; i < 10; ++i) {
    
            int cnt = counts[chars[i] - 'a'];
            for (int j = 0; j < words[i].size(); ++j) {
    
                counts[words[i][j] - 'a'] -= cnt;
            }
            while (cnt--) res += (nums[i] + '0');
        }
        sort(res.begin(), res.end());
        return res;
    }
};

class Solution {
    
public:
    string originalDigits(string s) {
    
        std::string res = "";
        vector<int> counts(128, 0), nums(10, 0);
        for (char c : s) {
    
            ++counts[c];
        }
        nums[0] = counts['z'];
        nums[2] = counts['w'];
        nums[4] = counts['u'];
        nums[6] = counts['x'];
        nums[8] = counts['g'];
        nums[1] = counts['o'] - nums[0] - nums[2] - nums[4];
        nums[3] = counts['h'] - nums[8];
        nums[5] = counts['f'] - nums[4];
        nums[7] = counts['s'] - nums[6];
        nums[9] = counts['i'] - nums[6] - nums[8] - nums[5];
        for (int i = 0; i < 10; ++i) {
    
            for (int j = 0; j < nums[i]; ++j) {
      //  There are several 1, There are several 2, There are several 3
                res += (i + '0');
            }
        }
        return res;
    }
};