zoukankan      html  css  js  c++  java
  • Leetcode: 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:
    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 }
  • 相关阅读:
    make 实例 一 3463
    python3 中对arrow库的总结(转发)
    impala 导出CSV 或excel
    设置虚拟机IP
    centos7 tomcat9
    eclipse 创建普通maven项目
    java log4j日志配置
    java运行jar命令提示没有主清单属性
    Java 读取 .properties 配置文件
    python 机器学习多项式回归
  • 原文地址:https://www.cnblogs.com/EdwardLiu/p/6132641.html
Copyright © 2011-2022 走看看