Go Java算法之从英文中重建数字示例详解

class Solution {
    static String[] ss = new String[]{"zero", "one", "two", "three", "four", "five", "six", "seven", "eight", "nine"};
    static int[] priority = new int[]{0, 8, 6, 3, 2, 7, 5, 9, 4, 1};
    public String originalDigits(String s) {
        int n = s.length();
        int[] cnts = new int[26];
        for (int i = 0; i < n; i++) cnts[s.charAt(i) - 'a']++;
        StringBuilder sb = new StringBuilder();
        for (int i : priority) {
            int k = Integer.MAX_VALUE;
            for (char c : ss[i].toCharArray()) k = Math.min(k, cnts[c - 'a']);
            for (char c : ss[i].toCharArray()) cnts[c - 'a'] -= k;
            while (k-- > 0) sb.append(i);
        }
        char[] cs = sb.toString().toCharArray();
        Arrays.sort(cs);
        return String.valueOf(cs);
    }
}

func originalDigits(s string) string {
    cnts, a := make([]int, 26), byte('a')
    for i := range s {
        cnts[s[i] - a]++
    }
    zeros, twos, fours, sixs, eights := cnts[byte('z') - a], cnts[byte('w') - a], cnts[byte('u') - a], cnts[byte('x') - a], cnts[byte('g') - a]
    fives, sevens, ones, threes := cnts[byte('f') - a] - fours, cnts[byte('s') - a] - sixs, cnts[byte('o') - a] - zeros - twos - fours, cnts[byte('h') - a] - eights
    nines := cnts[byte('i') - a] - fives - sixs - eights
    return strings.Repeat("0", zeros) + strings.Repeat("1", ones) + strings.Repeat("2", twos) + strings.Repeat("3", threes) + strings.Repeat("4", fours) + strings.Repeat("5", fives) + strings.Repeat("6", sixs) + strings.Repeat("7", sevens) + strings.Repeat("8", eights) + strings.Repeat("9", nines)
}