solve 17

Author: Zhang Xiaodong

src/solution/mod.rs

@@ -12,3 +12,4 @@ mod s0013_roman_to_integer;
 mod s0014_longest_common_prefix;
 mod s0015_3sum;
 mod s0016_3sum_closest;
+mod s0017_letter_combinations_of_a_phone_number;

src/solution/s0017_letter_combinations_of_a_phone_number.rs

@@ -0,0 +1,114 @@
+/**
+ * [17] 电话号码的字母组合
+ *
+ * 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。
+ * <img src="https://assets.leetcode-cn.com/aliyun-lc-upload/uploads/2021/11/09/200px-telephone-keypad2svg.png" style="width: 200px;" />
+ *  
+ * 示例 1：
+ *
+ * 输入：digits = "23"
+ * 输出：["ad","ae","af","bd","be","bf","cd","ce","cf"]
+ *
+ * 示例 2：
+ *
+ * 输入：digits = ""
+ * 输出：[]
+ *
+ * 示例 3：
+ *
+ * 输入：digits = "2"
+ * 输出：["a","b","c"]
+ *
+ *  
+ * 提示：
+ *
+ * 	0 <= digits.length <= 4
+ * 	digits[i] 是范围 ['2', '9'] 的一个数字。
+ *
+ */
+pub struct Solution {}
+
+// problem: https://leetcode.cn/problems/letter-combinations-of-a-phone-number/
+// discuss: https://leetcode.cn/problems/letter-combinations-of-a-phone-number/discuss/?currentPage=1&orderBy=most_votes&query=
+
+// submission codes start here
+
+impl Solution {
+    pub fn letter_combinations(digits: String) -> Vec<String> {
+        if digits.len() == 0 {
+            return vec![];
+        }
+        return Self::permutation(&digits);
+    }
+
+    fn permutation(digits: &str) -> Vec<String> {
+        if digits.len() == 1 {
+            let c = digits.as_bytes()[0];
+            if c == b'9' {
+                return vec![
+                    "w".to_string(),
+                    "x".to_string(),
+                    "y".to_string(),
+                    "z".to_string(),
+                ];
+            } else if c == b'7' {
+                return vec![
+                    "p".to_string(),
+                    "q".to_string(),
+                    "r".to_string(),
+                    "s".to_string(),
+                ];
+            } else if c == b'8' {
+                return vec!["t".to_string(), "u".to_string(), "v".to_string()];
+            }
+            let c = (c - b'2') * 3 + b'a';
+            return vec![
+                String::from_utf8(vec![c]).unwrap(),
+                String::from_utf8(vec![c + 1]).unwrap(),
+                String::from_utf8(vec![c + 2]).unwrap(),
+            ];
+        }
+        let vec1 = Self::permutation(&digits[0..1]);
+        let vec2 = Self::permutation(&digits[1..]);
+        let mut res = Vec::new();
+        for x in vec1.iter() {
+            for y in vec2.iter() {
+                res.push(format!("{}{}", x, y));
+            }
+        }
+        res
+    }
+}
+
+// submission codes end
+
+#[cfg(test)]
+mod tests {
+    use super::*;
+
+    #[test]
+    fn test_17() {
+        assert_eq!(
+            Solution::letter_combinations("2".to_string()),
+            vec!["a", "b", "c"]
+        );
+        assert_eq!(
+            Solution::letter_combinations("3".to_string()),
+            vec!["d", "e", "f"]
+        );
+        assert_eq!(
+            Solution::letter_combinations("9".to_string()),
+            vec!["w", "x", "y", "z"]
+        );
+        assert_eq!(
+            Solution::letter_combinations("23".to_string()),
+            vec!["ad", "ae", "af", "bd", "be", "bf", "cd", "ce", "cf"]
+        );
+        let emptyvec: Vec<String> = vec![];
+        assert_eq!(
+            Solution::letter_combinations("".to_string()),
+            emptyvec
+        );
+    }
+}