1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
121
122
123
124
125
126
127
128
129
130
131
132
133
134
135
136
137
138
139
140
141
142
143
144
145
146
147
148
149
150
151
152
153
154
155
156
157
158
159
160
161
162
163
164
165
166
167
168
169
170
171
172
173
174
175
176
177
178
179
180
181
182
183
184
185
186
187
188
189
190
191
192
193
194
195
//! # 问题描述
//!
//! 给定一个仅包含数字 `2-9` 的字符串,返回所有它能表示的字母组合。答案可以按**任意顺序**
//! 返回。
//! 
//! 给出数字到字母的映射如下(与电话按键相同)。注意 `1` 不对应任何字母。
//! 
//! ![](https://eastwind-cdn.dongs.xyz/image/20230224010445.png?w=256)
//! 
//! 示例 1:
//! 
//! ```plain
//! 输入:digits = "23"
//! 输出:["ad","ae","af","bd","be","bf","cd","ce","cf"]
//! ```
//! 
//! 示例 2:
//! 
//! ```plain
//! 输入:digits = ""
//! 输出:[]
//! ```
//! 
//! 示例 3:
//! 
//! ```plain
//! 输入:digits = "2"
//! 输出:["a","b","c"]
//! ```
//! 
//! 提示:
//! 
//! - `0 $\leqslant$ digits.length $\leqslant$ 4`
//! - `digits[i]` 是范围 `['2', '9']` 的一个数字。
//! 
//! 来源: <https://leetcode.cn/problems/letter-combinations-of-a-phone-number/>

////////////////////////////////////////////////////////////////////////////////

/// 电话号码的字母组合
///
/// # 参数
/// * `digits` - 输入的数字序列
///
/// # 示例
///
/// ```rust
/// use leetcode_rust::problems_cn::p000_0xx::p000_017::letter_combinations;
///
/// let input = String::from("2");
/// let output = letter_combinations(input);
/// assert_eq!(output, vec!["a", "b", "c"]);
/// ```
pub fn letter_combinations(digits: String) -> Vec<String> {
    // letter_combinations_by_match_control_flow(digits)
    letter_combinations_by_pre_allocation(digits)
    // letter_combinations_by_pre_allocation_v2(digits)
}

/// Match structure to get the letters for each digit
///
/// # Arguments
/// - `digits` - input digits
#[allow(dead_code)]
fn letter_combinations_by_match_control_flow(digits: String) -> Vec<String> {
    use std::str;
    let mut res: Vec<String> = vec![];
    for digit in digits.as_bytes() {
        let letters = match *digit {
            b'2' => "abc",
            b'3' => "def",
            b'4' => "ghi",
            b'5' => "jkl",
            b'6' => "mno",
            b'7' => "pqrs",
            b'8' => "tuv",
            b'9' => "wxyz",
            _ => "",
        }
        .as_bytes();

        if res.len() > 0 {
            let mut temp_res = vec![];
            for partial in res {
                for letter in letters {
                    let mut new_partial = partial.clone();
                    new_partial.push(*letter as char);
                    temp_res.push(new_partial);
                }
            }
            res = temp_res;
        } else {
            res = letters
                .iter()
                .map(|l| str::from_utf8(vec![*l].as_slice()).unwrap().to_string())
                .collect();
        }
    }

    res
}

/// Create a vector of bytes representing each digit in advance. Save memory usage
/// significantly.
///
/// # Arguments
/// - `digits` - input digits
#[allow(dead_code)]
fn letter_combinations_by_pre_allocation(digits: String) -> Vec<String> {
    use std::str;

    let digits_to_letters: Vec<&[u8]> =
        vec!["abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"]
            .iter()
            .map(|s| s.as_bytes())
            .collect();

    let mut res: Vec<String> = vec![];
    for digit in digits.as_bytes() {
        let letters = digits_to_letters[(*digit - b'2') as usize];

        if res.len() > 0 {
            let mut temp_res = vec![];
            for partial in res {
                for letter in letters {
                    let mut new_partial = partial.clone();
                    new_partial.push(*letter as char);
                    temp_res.push(new_partial);
                }
            }
            res = temp_res;
        } else {
            let mut temp: Vec<String> = letters
                .iter()
                .map(|l| str::from_utf8(vec![*l].as_slice()).unwrap().to_string())
                .collect();
            res.append(&mut temp);
        }
    }

    res
}

/// Create a vector of bytes representing each digit in advance. Save memory
/// usage significantly. Simplified expression but may cause a little more
/// memory consumption.
///
/// # Arguments
/// - `digits` - input digits
#[allow(dead_code)]
fn letter_combinations_by_pre_allocation_v2(digits: String) -> Vec<String> {
    use std::str;

    let digits_to_letters: Vec<&[u8]> =
        vec!["abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"]
            .iter()
            .map(|s| s.as_bytes())
            .collect();

    let mut res: Vec<String> = vec![];
    for digit in digits.as_bytes() {
        let letters = digits_to_letters[(*digit - b'2') as usize];

        res = if res.len() == 0 {
            letters
                .iter()
                .map(|l| str::from_utf8(&[*l]).unwrap().to_string())
                .collect()
        } else {
            let mut temp_res = vec![];
            for partial in res {
                for letter in letters {
                    temp_res.push(partial.clone() + str::from_utf8(&[*letter]).unwrap());
                }
            }
            temp_res
        };
    }

    res
}

#[cfg(test)]
mod tests {
    use super::letter_combinations;
    #[test]
    fn test_letter_combinations() {
        assert_eq!(
            letter_combinations(String::from("23")),
            vec!["ad", "ae", "af", "bd", "be", "bf", "cd", "ce", "cf"]
        );
        assert_eq!(letter_combinations(String::from("")), Vec::<String>::new());
        assert_eq!(letter_combinations(String::from("2")), vec!["a", "b", "c"]);
    }
}