423. Reconstruct Original Digits from English

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分类专栏： LeetCode Math 文章标签： LeetCode Math Reconstruct Original

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

Note:

Input contains only lowercase English letters.
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.
Input length is less than 50,000.

Example 1:

Input: "owoztneoer"

Output: "012"

Example 2:

Input: "fviefuro"

Output: "45"

根据每个数中英文字母的特殊性，统计数的个数，最后遍历打印。代码如下：

public class Solution {
    public String originalDigits(String s) {
        int[] count = new int[10];
        for (char ch: s.toCharArray()){
            if(ch == 'z') count[0]++;
            else if (ch == 'o') count[1]++;
            else if (ch == 'w') count[2]++;
            else if (ch == 'r') count[3]++;
            else if (ch == 'u') count[4]++;
            else if (ch == 'f') count[5]++;
            else if (ch == 'x') count[6]++;
            else if (ch == 's') count[7]++;
            else if (ch == 'g') count[8]++;
            else if (ch == 'i') count[9]++;
        }
        count[1] = count[1] - count[0] - count[2] - count[4];
        count[3] = count[3] - count[0] - count[4];
        count[5] = count[5] - count[4];
        count[7] = count[7] - count[6];
        count[9] = count[9] - count[8] - count[6] - count[5];
        StringBuilder sb = new StringBuilder();
        for (int i = 0; i < 10; i ++) {
            for (int j = 0; j < count[i]; j ++) {
                sb.append(i);
            }
        }
        return sb.toString();
    }
}