Today is an example of The non-Designer's Design Book on Page 54 and Check If a Word Occurs As a Prefix of Any Word in a Sentence on leetcode with cpp.

The non-Designer's Design Book

my answer:

• [T] color scheme of top and bottom
• [F] heart shape
• [F] subtitle's font and color
• [T] dividing line
• [T] font of title and subtitle

Reconstruct Original Digits from English

Description

Given a non-empty string containing an out-of-order English representation of digits 0-9, output the digits in ascending order.

Note:

1. Input contains only lowercase English letters.
2. Input is guaranteed to be valid and can be transformed to its original digits. That means invalid inputs such as "abc" or "zerone" are not permitted.
3. Input length is less than 50,000.

Example 1:

Input: "owoztneoer"

Output: "012"

Example 2:

Input: "fviefuro"

Output: "45"

Solution

there are unique characters in words so that you can easily find what numbers in the string.

class Solution {
  int cnt[256];
public:
  string originalDigits(string s) {
    for(auto &a: s) {
      cnt[a] += 1;
    }
    while(cnt['z']) {
      cnt['z'] -= 1;
      cnt['e'] -= 1;
      cnt['r'] -= 1;
      cnt['o'] -= 1;
      cnt[0] += 1;
    }
    while(cnt['x']) {
      cnt['s'] -= 1;
      cnt['i'] -= 1;
      cnt['x'] -= 1;
      cnt[6] += 1;
    }
    while(cnt['w']) {
      cnt['t'] -= 1;
      cnt['w'] -= 1;
      cnt['o'] -= 1;
      cnt[2] += 1;
    }
    while(cnt['g']) {
      cnt['e'] -= 1;
      cnt['i'] -= 1;
      cnt['g'] -= 1;
      cnt['h'] -= 1;
      cnt['t'] -= 1;
      cnt[8] += 1;
    }
    while(cnt['s']) {
      cnt['s'] -= 1;
      cnt['e'] -= 1;
      cnt['v'] -= 1;
      cnt['e'] -= 1;
      cnt['n'] -= 1;
      cnt[7] += 1;
    }
    while(cnt['v']) {
      cnt['f'] -= 1;
      cnt['i'] -= 1;
      cnt['v'] -= 1;
      cnt['e'] -= 1;
      cnt[5] += 1;
    }
    while(cnt['f']) {
      cnt['f'] -= 1;
      cnt['o'] -= 1;
      cnt['u'] -= 1;
      cnt['r'] -= 1;
      cnt[4] += 1;
    }
    while(cnt['t']) {
      cnt['t'] -= 1;
      cnt['h'] -= 1;
      cnt['r'] -= 1;
      cnt['e'] -= 1;
      cnt['e'] -= 1;
      cnt[3] += 1;
    }
    while(cnt['i']) {
      cnt['n'] -= 1;
      cnt['i'] -= 1;
      cnt['n'] -= 1;
      cnt['e'] -= 1;
      cnt[9] += 1;
    }
    while(cnt['o']) {
      cnt['o'] -= 1;
      cnt['n'] -= 1;
      cnt['e'] -= 1;
      cnt[1] += 1;
    }
    stringstream ss;
    for(int i = 0; i < 10; ++i) {
      while(cnt[i]) {
        ss << i;
        cnt[i] -= 1;
      }
    }
    return ss.str();
  }
};