小米面试题：
手机分身，电话号码隐藏。
0-9分别对应ZERO,ONE,TWO,...,NINE
为保证手机拨号安全性，拨号后，存为对应的字符串，并打乱顺序，保证安全性。
现在给定一个字符串，求最小的数字组合
例：
TWOONE
OTNWOE
12
12
分析：
| 关键字搜索： | 第一次 | 第二次 | 第三次 |
| 0 ZERO | Z | ||
| 1 ONE | O | ||
| 2 TWO | W | ||
| 3 THREE | R | ||
| 4 FOUR | U | ||
| 5 FIVE | F | ||
| 6 SIX | X | ||
| 7 SENVE | S | ||
| 8 EIGHT | G | ||
| 9 NINE | I/E |
代码实现如下：
1 #include "stdafx.h"
2 #include<string>
3 #include<iostream>
4 using namespace std;
5
6 int main()
7 {
8 int zidian[26],number[10];
9 memset(zidian,0,26*sizeof(int));
10 memset(number,0,10*sizeof(int));
11 string numberString;
12 while(cin >> numberString)
13 {//while循环，无限获取测试用例
14 for (int i = 0; i < numberString.size(); i++)
15 {//for循环建立字典索引，直接针对字典查，而不再单个搜索。
16 char tempChar = numberString[i];
17 zidian[(int)tempChar - 65] += 1;
18 }
19 int tempNum;
20 //查找0 ZERO Z
21 if(zidian[(int)‘Z‘ - 65] != 0)
22 {
23 tempNum = zidian[(int)‘Z‘ - 65];
24 zidian[(int)‘Z‘ - 65] = 0;
25 zidian[(int)‘E‘ - 65] -= tempNum;
26 zidian[(int)‘R‘ - 65] -= tempNum;
27 zidian[(int)‘O‘ - 65] -= tempNum;
28 number[0] = tempNum;
29 }
30 //查找2 TWO W
31 if(zidian[(int)‘W‘ - 65] != 0)
32 {
33 tempNum = zidian[(int)‘W‘ - 65];
34 zidian[(int)‘W‘ - 65] = 0;
35 zidian[(int)‘T‘ - 65] -= tempNum;
36 zidian[(int)‘O‘ - 65] -= tempNum;
37 number[2] = tempNum;
38 }
39 //查找4 FOUR U
40 if(zidian[(int)‘U‘ - 65] != 0)
41 {
42 tempNum = zidian[(int)‘U‘ - 65];
43 zidian[(int)‘U‘ - 65] = 0;
44 zidian[(int)‘F‘ - 65] -= tempNum;
45 zidian[(int)‘R‘ - 65] -= tempNum;
46 zidian[(int)‘O‘ - 65] -= tempNum;
47 number[4] = tempNum;
48 }
49 //查找6 SIX X
50 if(zidian[(int)‘X‘ - 65] != 0)
51 {
52 tempNum = zidian[(int)‘X‘ - 65];
53 zidian[(int)‘X‘ - 65] = 0;
54 zidian[(int)‘S‘ - 65] -= tempNum;
55 zidian[(int)‘I‘ - 65] -= tempNum;
56 number[6] = tempNum;
57 }
58 //查找8 EIGHT G
59 if(zidian[(int)‘G‘ - 65] != 0)
60 {
61 tempNum = zidian[(int)‘G‘ - 65];
62 zidian[(int)‘G‘ - 65] = 0;
63 zidian[(int)‘E‘ - 65] -= tempNum;
64 zidian[(int)‘I‘ - 65] -= tempNum;
65 zidian[(int)‘H‘ - 65] -= tempNum;
66 zidian[(int)‘T‘ - 65] -= tempNum;
67 number[8] = tempNum;
68 }
69 //查找1 ONE O
70 if(zidian[(int)‘O‘ - 65] != 0)
71 {
72 tempNum = zidian[(int)‘O‘ - 65];
73 zidian[(int)‘O‘ - 65] = 0;
74 zidian[(int)‘N‘ - 65] -= tempNum;
75 zidian[(int)‘E‘ - 65] -= tempNum;
76 number[1] = tempNum;
77 }
78 //查找3 THREE R
79 if(zidian[(int)‘R‘ - 65] != 0)
80 {
81 tempNum = zidian[(int)‘R‘ - 65];
82 zidian[(int)‘R‘ - 65] = 0;
83 zidian[(int)‘T‘ - 65] -= tempNum;
84 zidian[(int)‘H‘ - 65] -= tempNum;
85 zidian[(int)‘E‘ - 65] -= 2 * tempNum;
86 number[3] = tempNum;
87 }
88 //查找5 FIVE F
89 if(zidian[(int)‘F‘ - 65] != 0)
90 {
91 tempNum = zidian[(int)‘F‘ - 65];
92 zidian[(int)‘F‘ - 65] = 0;
93 zidian[(int)‘I‘ - 65] -= tempNum;
94 zidian[(int)‘V‘ - 65] -= tempNum;
95 zidian[(int)‘E‘ - 65] -= tempNum;
96 number[5] = tempNum;
97 }
98 //查找7 SEVEN S
99 if(zidian[(int)‘S‘ - 65] != 0)
100 {
101 tempNum = zidian[(int)‘S‘ - 65];
102 zidian[(int)‘S‘ - 65] = 0;
103 zidian[(int)‘E‘ - 65] -= 2 * tempNum;
104 zidian[(int)‘V‘ - 65] -= tempNum;
105 zidian[(int)‘N‘ - 65] -= tempNum;
106 number[4] = tempNum;
107 }
108 //查找9 NINE I
109 if(zidian[(int)‘I‘ - 65] != 0)
110 {
111 tempNum = zidian[(int)‘I‘ - 65];
112 zidian[(int)‘I‘ - 65] = 0;
113 zidian[(int)‘N‘ - 65] -= 2 * tempNum;
114 zidian[(int)‘E‘ - 65] -= tempNum;
115 number[4] = tempNum;
116 }
117 //转化为字符串输出（保证最小）
118 string sum;
119 for (int i = 0; i < 10; i++)
120 {
121 string temps(number[i],(char)(i+48));
122 sum += temps;
123 }
124 cout << sum << endl;
125 }
126 return 0;
127 }
C++ Code
时间： 2024-08-12 06:29:10