Solve #258

Author: aylei

src/lib.rs

@@ -206,3 +206,5 @@ mod n0239_sliding_window_maximum;
 mod n0241_different_ways_to_add_parentheses;
 mod n0242_valid_anagram;
 mod n0257_binary_tree_paths;
+mod n0258_add_digits;
+mod n0260_single_number_iii;

src/n0258_add_digits.rs

@@ -0,0 +1,38 @@
+/**
+ * [258] Add Digits
+ *
+ * Given a non-negative integer num, repeatedly add all its digits until the result has only one digit.
+ *
+ * Example:
+ *
+ *
+ * Input: 38
+ * Output: 2
+ * Explanation: The process is like: 3 + 8 = 11, 1 + 1 = 2.
+ *              Since 2 has only one digit, return it.
+ *
+ *
+ * Follow up:<br />
+ * Could you do it without any loop/recursion in O(1) runtime?
+ */
+pub struct Solution {}
+
+// submission codes start here
+
+impl Solution {
+    pub fn add_digits(num: i32) -> i32 {
+        1 + ((num - 1) % 9)
+    }
+}
+
+// submission codes end
+
+#[cfg(test)]
+mod tests {
+    use super::*;
+
+    #[test]
+    fn test_258() {
+        assert_eq!(Solution::add_digits(1234), 1);
+    }
+}

src/n0260_single_number_iii.rs

@@ -0,0 +1,37 @@
+/**
+ * [260] Single Number III
+ *
+ * Given an array of numbers nums, in which exactly two elements appear only once and all the other elements appear exactly twice. Find the two elements that appear only once.
+ *
+ * Example:
+ *
+ *
+ * Input:  [1,2,1,3,2,5]
+ * Output: [3,5]
+ *
+ * Note:
+ *
+ * <ol>
+ * 	The order of the result is not important. So in the above example, [5, 3] is also correct.
+ * 	Your algorithm should run in linear runtime complexity. Could you implement it using only constant space complexity?
+ * </ol>
+ */
+pub struct Solution {}
+
+// submission codes start here
+
+impl Solution {
+    pub fn single_number(nums: Vec<i32>) -> Vec<i32> {
+        vec![]
+    }
+}
+
+// submission codes end
+
+#[cfg(test)]
+mod tests {
+    use super::*;
+
+    #[test]
+    fn test_260() {}
+}