423.Reconstruct Original Digits from English

423.Reconstruct Original Digits from English

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"

/*解题思路
423.Reconstruct Original Digits from English
对字符串中的字符进行计数，根据这些字符与数字的关系可以直接得到结果。
例如：字符z只在zero中出现，字符w只在two中出现，字符x只在six中出现，
字符g只在eigth中出现，字符u只在four中出现那么我们根据z,w,x,g,u的个
数就可以知道0,2,6,8,4的个数。对于剩下的one,three,five,seven，one可
以由字符o的个数减去在0,2,4,6,8中出现的o的个数。three可以由字符h的个
数减去字符t,r,e在0,2,4,6,8中出现的个数同理可以得到剩下的数字的个数。
*/
public class Solution{
	public static void main(String[] args){
		//TODO auto-generated method stub
		System.out.println(Solution.originalDigits("owoztneoer"));
	}
	//找到0-9独有的字母
	public static String originalDigits(String s){
			if(s == null || s.length() == 0)
				return NULL;
			int[] count = new int[10]; //存放0-9出现的次数
			for(char c:s.toCharArray())
			{
				if(c == 'z') count[0]++;
				if(c == 'w') count[2]++;
				if(c == 'x') count[6]++;
				if(c == 's') count[7]++;
				if(c == 'g') count[8]++;
				if(c == 'u') count[4]++;
				if(c == 'f') count[5]++;
				if(c == 'h') count[3]++;
				if(c == 'i') count[9]++;
				if(c == 'o') count[1]++;
			}
			count[7] -= count[6];
			count[5] -= count[4];
			count[3] -= count[8];
			count[9] = count[9] -count[5] -count[6] -count[8];
			count[1] = count[1] -count[2] -count[0] -count[4];
			StringBuilder sb = new StringBuilder();
			for(int i = 0; i<10; ++i)
			{
				for(int j = 1; j <= count[i]; ++j)
				{
					sb.append9(i);
				}
			return sb.toString();
			}
	}
}