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"
思路一：
//1. 循环一遍，统计所有字母各自总数量
//2.对 zero two four six eight 统计（因为10个数字中，它们有各自独特的标记，分别是 z w u x g）,出现标记一次，统计总数对相应的字母减1，如出现z，则zero4个字母都减去1
//3.对剩下的数中，继续找独特点， 分别有 one three five seven （标记为 o t f s），统计总数对相应的字母减1
//4.对剩下的nine 进行统计，i或者e出现几次，就有几个nine
// z one w three u five x seven g nine for [ a b c.... ] 比如出现z就 zero 都-1
代码如下：
import java.util.HashMap;
public class Solution {
static HashMap<String, Integer> hashMap;
public static void replace(String str)
{
hashMap.replace(str, hashMap.get(str)-1);
}
public String originalDigits(String s) {
int [] ans=new int[10];
hashMap= new HashMap<String, Integer>();
String str=null;
for(int i=0;i<26;i++)
{
str=String.valueOf((char)(i+97));
hashMap.put(str,0);
}
for (int i = 0; i < s.length(); i++) {
str=s.substring(i, i+1);
hashMap.replace(str, hashMap.get(str)+1);
}
while(hashMap.get("z")>0)
{
replace("z");
replace("e");
replace("r");
replace("o");
ans[0]+=1;
}
while(hashMap.get("w")>0)
{
replace("t");
replace("w");
replace("o");
ans[2]+=1;
}
while(hashMap.get("u")>0)
{
replace("f");
replace("o");
replace("u");
replace("r");
ans[4]+=1;
}
while(hashMap.get("x")>0)
{
replace("s");
replace("i");
replace("x");
ans[6]+=1;
}
while(hashMap.get("g")>0)
{
replace("e");
replace("i");
replace("g");
replace("h");
replace("t");
ans[8]+=1;
}
while(hashMap.get("o")>0)
{
replace("o");
replace("n");
replace("e");
ans[1]+=1;
}
while(hashMap.get("t")>0)
{
replace("t");
replace("h");
replace("r");
replace("e");
replace("e");
ans[3]+=1;
}
while(hashMap.get("f")>0)
{
replace("f");
replace("i");
replace("v");
replace("e");
ans[5]+=1;
}
while(hashMap.get("s")>0)
{
replace("s");
replace("e");
replace("v");
replace("e");
replace("n");
ans[7]+=1;
}
while(hashMap.get("i")>0)
{
replace("n");
replace("i");
replace("n");
replace("e");
ans[9]+=1;
}
StringBuilder sb = new StringBuilder();
for (int i = 0; i <= 9; i++){
for (int j = 0; j < count[i]; j++){
sb.append(i);
}
}
return sb.toString();
}
}
但是以上代码比较冗长，把思路一转换一下，先对所有标记字符计数，再用总数减去相应的数量，得到一个正确的答案，就可以很简短的写出来，代码很容易理解
代码如下：
public String originalDigits(String s) {
int[] count = new int[10];
for (int i = 0; i < s.length(); i++){
if (c == 'z') count[0]++;
if (c == 'w') count[2]++;
if (c == 'x') count[6]++;
if (c == 'g') count[8]++;
if (c == 'u') count[4]++;
if (c == 's') count[7]++;
if (c == 'f') count[5]++;
if (c == 'h') count[3]++;
if (c == 'i') count[9]++;
if (c == 'o') count[1]++;
}
count[7] -= count[6];//（six，seven都有s，那么s的总数量减去6的数量就是7的数量），下面同理
count[5] -= count[4];
count[3] -= count[8];
count[9] = count[9] - count[8] - count[5] - count[6];
count[1] = count[1] - count[0] - count[2] - count[4];
StringBuilder sb = new StringBuilder();
for (int i = 0; i <= 9; i++){
for (int j = 0; j < count[i]; j++){
sb.append(i);
}
}
return sb.toString();
}