Leetcode 423. Reconstruct Original Digits from English

题意

给你字符串，保证这些字符串能组合成0-9的英文单词，如zero,one,two...，请你按顺序输出这些数字，如ezowrte，能组合成zero,two，那就输出02

解答

我发现zero，one，two...nine,在这些单词里，有些字母是只出现在一个单词里的，比如z出现在zero，那么我就能通过z来得到zero出现的次数，所以我就统计所有字母出现的次数来计算每个单词出现次数

public class Solution {
    public String originalDigits(String s) {
        StringBuilder sb = new StringBuilder();
        int[] cnt = new int[26];
        int[] num = new int[10];
        for (int i = 0; i < s.length(); i++) {
            cnt[s.charAt(i)-'a']++;
        }
        int x;
        // zero
        x = cnt['z'-'a'];
        num[0] = x;
        cnt['z'-'a'] -= x;
        cnt['e'-'a'] -= x;
        cnt['r'-'a'] -= x;
        cnt['o'-'a'] -= x;
        
        // two
        x = cnt['w'-'a'];
        num[2] = x;
        cnt['t'-'a'] -= x;
        cnt['w'-'a'] -= x;
        cnt['o'-'a'] -= x;
        
        
        // four
        x = cnt['u'-'a'];
        num[4] = x;
        cnt['f'-'a'] -= x;
        cnt['o'-'a'] -= x;
        cnt['u'-'a'] -= x;
        cnt['r'-'a'] -= x;
        
        // six
        x = cnt['x'-'a'];
        num[6] = x;
        cnt['s'-'a'] -= x;
        cnt['i'-'a'] -= x;
        cnt['x'-'a'] -= x;
        
        // one
        x = cnt['o'-'a'];
        num[1] = x;
        cnt['o'-'a'] -= x;
        cnt['n'-'a'] -= x;
        cnt['e'-'a'] -= x;
        
        // three
        x = cnt['r'-'a'];
        num[3] = x;
        cnt['t'-'a'] -= x;
        cnt['h'-'a'] -= x;
        cnt['r'-'a'] -= x;
        cnt['e'-'a'] -= x;
        cnt['e'-'a'] -= x;
        
        // five
        x = cnt['f'-'a'];
        num[5] = x;
        cnt['f'-'a'] -= x;
        cnt['i'-'a'] -= x;
        cnt['v'-'a'] -= x;
        cnt['e'-'a'] -= x;
        
        // seven
        x = cnt['v'-'a'];
        num[7] = x;
        cnt['s'-'a'] -= x;
        cnt['e'-'a'] -= x;
        cnt['v'-'a'] -= x;
        cnt['e'-'a'] -= x;
        cnt['n'-'a'] -= x;
        
        // nine
        x = cnt['n'-'a'];
        x/=2;
        num[9] = x;
        cnt['n'-'a'] -= x;
        cnt['i'-'a'] -= x;
        cnt['n'-'a'] -= x;
        cnt['e'-'a'] -= x;
        
        
        // eight
        x = cnt['e'-'a'];
        num[8] = x;
        cnt['e'-'a'] -= x;
        cnt['i'-'a'] -= x;
        cnt['g'-'a'] -= x;
        cnt['h'-'a'] -= x;
        cnt['t'-'a'] -= x;
        
        for (int i = 0; i < 10; i++) {
            while (num[i] > 0) {
                sb.append(""+i);
                num[i]--;
            }
        }
        return sb.toString();
    }
}