solve #58

Author: aylei

src/lib.rs

@@ -58,3 +58,4 @@ mod n0054_spiral_matrix;
 mod n0055_jump_game;
 mod n0056_merge_intervals;
 mod n0057_insert_interval;
+mod n0058_length_of_last_word;

src/n0058_length_of_last_word.rs

@@ -0,0 +1,52 @@
+/**
+ * [58] Length of Last Word
+ *
+ * Given a string s consists of upper/lower-case alphabets and empty space characters ' ', return the length of last word in the string.
+ * 
+ * If the last word does not exist, return 0.
+ * 
+ * Note: A word is defined as a character sequence consists of non-space characters only.
+ * 
+ * Example:
+ * 
+ * Input: "Hello World"
+ * Output: 5
+ * 
+ * 
+ */
+pub struct Solution {}
+
+// submission codes start here
+
+impl Solution {
+    pub fn length_of_last_word(s: String) -> i32 {
+        let seq: Vec<char> = s.chars().rev().collect();
+        let mut result = 0;
+        let mut find = false;
+        for ch in seq {
+            if ch == ' ' && find {
+                break;
+            }
+            if ch != ' ' {
+                find = true;
+                result += 1;
+            }
+        }
+        result
+    }
+}
+
+// submission codes end
+
+#[cfg(test)]
+mod tests {
+    use super::*;
+
+    #[test]
+    fn test_58() {
+        assert_eq!(Solution::length_of_last_word("Hello World".to_owned()), 5);
+        assert_eq!(Solution::length_of_last_word("       ".to_owned()), 0);
+        assert_eq!(Solution::length_of_last_word("".to_owned()), 0);
+        assert_eq!(Solution::length_of_last_word("     rrrrr  ".to_owned()), 5);
+    }
+}