423. Reconstruct Original Digits from English

转自：https://www.cnblogs.com/charlesblc/p/5967694.html

public class Solution {    // zero one two three four five six seven eight nine ten    // z 0    // e 0 1 3 3 5 7 7 8 9    // r 0 3 4    // o 0 1 2 4    // n 1 7 9 9    // t 2 3 8    // w 2    // h 3 8    // f 4 5    // u 4    // i 5 6 8 9    // v 5 7    // s 6 7    // x 6    // g 8    public String originalDigits(String s) {        // 太牛了,开始我也想到统计,但是想到删除字符串那些复杂的操作上面去了        int[] counts = new int[10];        for (int i=0; i<s.length(); i++) {            char ch = s.charAt(i);            if (ch == 'z') counts[0]++;            if (ch == 'u') counts[4]++;            if (ch == 'w') counts[2]++;            if (ch == 'x') counts[6]++;            if (ch == 'g') counts[8]++;            if (ch == 'h') counts[3]++; // 3, 8            if (ch == 's') counts[7]++; // 6, 7            if (ch == 'f') counts[5]++; // 4, 5            if (ch == 'o') counts[1]++; // 0, 1, 2, 4            if (ch == 'i') counts[9]++; // 5, 6, 8, 9        }        counts[3] -= counts[8];        counts[7] -= counts[6];        counts[5] -= counts[4];        counts[1] -= counts[0] + counts[2] + counts[4];        counts[9] -= counts[5] + counts[6] + counts[8];        StringBuilder sb = new StringBuilder();        for (int i=0; i<10; i++) {            for (int j=0; j<counts[i]; j++) {                sb.append(i);            }        }        return sb.toString();    }}