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[LeetCode]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"

题意:给出一串乱序的数字单词,输出从小到大排序的对应数字

思路:

  利用暗藏的信息,找出每个数字的英文单词对应的特别的字母,比如two中的w,four中的u,eight中的g,six的x,zero的z,都是唯一的

  three的h,five的f,seven的s出现了两次

  one中的o出现了四次

  nine中的i出现了三次

利用这些特点,把这些字母作为所对应数字的唯一标示进行统计,之后进行数学处理,比如three中的h同时再eight中出现了,那么three的数量减去

eight的数量,就是three的真实数量

C代码如下:

 1 char* originalDigits(char* s) {
 2     int count[10]={0};
 3     int i=0,j=0;
 4     char *str=malloc(sizeof(char)*16000); //注意只能返回栈空间的变量
 5     while(s[i])
 6     {
 7         if(z==s[i]) count[0]++;
 8         else if (w==s[i]) count[2]++;
 9         else if (u==s[i]) count[4]++;
10         else if (x==s[i]) count[6]++;
11         else if (g==s[i]) count[8]++;
12         else if (f==s[i]) count[5]++;
13         else if (h==s[i]) count[3]++;
14         else if (s==s[i]) count[7]++;
15         else if (i==s[i]) count[9]++;
16         else if (o==s[i]) count[1]++;
17         i++;
18     }
19     count[5]-=count[4];
20     count[3]-=count[8];
21     count[7]-=count[6];
22     count[1]-=(count[2]+count[4]+count[0]);
23     count[9]-=(count[5]+count[6]+count[8]);
24     for(i=0;i<10;i++)
25         {
26             if(count[i])
27                 while(count[i]--)
28                 {
29                     str[j]=0+i;
30                     j++;
31                 }
32         }
33     str[j]=\0;
34     return str;
35 }

 

[LeetCode]423. Reconstruct Original Digits from English