码迷,mamicode.com
首页 > 其他好文 > 详细

Leetcode: Reconstruct Original Digits from English

时间:2016-12-05 09:30:45      阅读:251      评论:0      收藏:0      [点我收藏+]

标签:case   char   original   ted   tput   null   color   end   transform   

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"

# of ‘0‘: # of ‘z‘

# of ‘2‘: # of ‘w‘

4: u

6: x

8: g

3: h - 8

5: f - 4

7: s - 6

1: o - 0 - 2 - 4

9: i - 5 - 6 - 8

 1 public String originalDigits(String s) {
 2     int[] count = new int[10];
 3     for (int i = 0; i < s.length(); i++){
 4         char c = s.charAt(i);
 5         if (c == ‘z‘) count[0]++;
 6         if (c == ‘w‘) count[2]++;
 7         if (c == ‘x‘) count[6]++;
 8         if (c == ‘s‘) count[7]++; //7-6
 9         if (c == ‘g‘) count[8]++;
10         if (c == ‘u‘) count[4]++; 
11         if (c == ‘f‘) count[5]++; //5-4
12         if (c == ‘h‘) count[3]++; //3-8
13         if (c == ‘i‘) count[9]++; //9-8-5-6
14         if (c == ‘o‘) count[1]++; //1-0-2-4
15     }
16     count[7] -= count[6];
17     count[5] -= count[4];
18     count[3] -= count[8];
19     count[9] = count[9] - count[8] - count[5] - count[6];
20     count[1] = count[1] - count[0] - count[2] - count[4];
21     StringBuilder sb = new StringBuilder();
22     for (int i = 0; i <= 9; i++){
23         for (int j = 0; j < count[i]; j++){
24             sb.append(i);
25         }
26     }
27     return sb.toString();
28 }

 

我的code用了一个数组来存char count

 1 public class Solution {
 2     public String originalDigits(String s) {
 3         StringBuilder res = new StringBuilder();
 4         if (s==null || s.length()==0) return "";
 5         int[] chars = new int[26];
 6         int[] digits = new int[10];
 7         for (int i=0; i<s.length(); i++) {
 8             chars[s.charAt(i)-‘a‘]++;
 9         }
10         count(chars, digits);
11         for (int i=0; i<digits.length; i++) {
12             for (int j=0; j<digits[i]; j++) {
13                 res.append(i);
14             }
15         }
16         return res.toString();
17     } 
18     
19     public void count(int[] chars, int[] digits) {
20         //‘0‘
21         digits[0] = chars[‘z‘-‘a‘];
22         //‘2‘
23         digits[2] = chars[‘w‘-‘a‘];
24         //‘4‘
25         digits[4] = chars[‘u‘-‘a‘];
26         //‘6‘
27         digits[6] = chars[‘x‘-‘a‘];
28         //‘8‘
29         digits[8] = chars[‘g‘-‘a‘];
30         //‘1‘ and ‘2‘ and ‘0‘ and ‘4‘ share ‘o‘
31         digits[1] = chars[‘o‘-‘a‘] - digits[2] - digits[0] - digits[4];
32         //‘3‘ and ‘8‘ share ‘h‘
33         digits[3] = chars[‘h‘-‘a‘] - digits[8];
34         //‘5‘ and ‘4‘ share ‘f‘
35         digits[5] = chars[‘f‘-‘a‘] - digits[4];
36         //‘7‘ and ‘6‘ share ‘s‘
37         digits[7] = chars[‘s‘-‘a‘] - digits[6];
38         //‘9‘ and ‘5‘ and ‘6‘ and ‘8‘ share ‘i‘
39         digits[9] = chars[‘i‘-‘a‘] - digits[5] - digits[6] - digits[8];
40     }
41 }

 

Leetcode: Reconstruct Original Digits from English

标签:case   char   original   ted   tput   null   color   end   transform   

原文地址:http://www.cnblogs.com/EdwardLiu/p/6132641.html

(0)
(0)
   
举报
评论 一句话评论(0
登录后才能评论!
© 2014 mamicode.com 版权所有  联系我们:gaon5@hotmail.com
迷上了代码!