diff --git a/src/solution/mod.rs b/src/solution/mod.rs
index abf3f89..c441aad 100644
--- a/src/solution/mod.rs
+++ b/src/solution/mod.rs
@@ -8,3 +8,4 @@ mod s0162_find_peak_element;
 mod s2828_check_if_a_string_is_an_acronym_of_words;
 mod s0052_n_queens_ii;
 mod s1276_number_of_burgers_with_no_waste_of_ingredients;
+mod s0006_zigzag_conversion;
diff --git a/src/solution/s0006_zigzag_conversion.rs b/src/solution/s0006_zigzag_conversion.rs
new file mode 100644
index 0000000..510085d
--- /dev/null
+++ b/src/solution/s0006_zigzag_conversion.rs
@@ -0,0 +1,98 @@
+/**
+ * [6] Zigzag Conversion
+ *
+ * The string "PAYPALISHIRING" is written in a zigzag pattern on a given number of rows like this: (you may want to display this pattern in a fixed font for better legibility)
+ * 
+ * P   A   H   N
+ * A P L S I I G
+ * Y   I   R
+ * 
+ * And then read line by line: "PAHNAPLSIIGYIR"
+ * Write the code that will take a string and make this conversion given a number of rows:
+ * 
+ * string convert(string s, int numRows);
+ * 
+ *  
+ * <strong class="example">Example 1:
+ * 
+ * Input: s = "PAYPALISHIRING", numRows = 3
+ * Output: "PAHNAPLSIIGYIR"
+ * 
+ * <strong class="example">Example 2:
+ * 
+ * Input: s = "PAYPALISHIRING", numRows = 4
+ * Output: "PINALSIGYAHRPI"
+ * Explanation:
+ * P     I    N
+ * A   L S  I G
+ * Y A   H R
+ * P     I
+ * 
+ * <strong class="example">Example 3:
+ * 
+ * Input: s = "A", numRows = 1
+ * Output: "A"
+ * 
+ *  
+ * Constraints:
+ * 
+ * 	1 <= s.length <= 1000
+ * 	s consists of English letters (lower-case and upper-case), ',' and '.'.
+ * 	1 <= numRows <= 1000
+ * 
+ */
+pub struct Solution {}
+
+// problem: https://leetcode.cn/problems/zigzag-conversion/
+// discuss: https://leetcode.cn/problems/zigzag-conversion/discuss/?currentPage=1&orderBy=most_votes&query=
+
+// submission codes start here
+
+impl Solution {
+    pub fn convert(s: String, num_rows: i32) -> String {
+        let length = s.len() as i32;
+        if num_rows < 2 {
+            return s;
+        }
+
+        let mut result = String::new();
+        let s: Vec<char> = s.chars().collect();
+
+        for row in 0..num_rows {
+            for begin in (0..length).step_by((2 * num_rows - 2) as usize) {
+                // 竖线上的元素
+                let i = begin + row;
+
+                if i >= length {
+                    break;
+                }
+                result.push(s[i as usize]);
+
+                // 斜线上的元素
+                let j = i + 2 * (num_rows - row - 1);
+                if j >= length {
+                    break;
+                } else if j == i + 2 * (num_rows - 1) || i == j {
+                    continue;
+                }
+                result.push(s[j as usize]);
+            }
+        }
+
+        result
+    }
+}
+
+// submission codes end
+
+#[cfg(test)]
+mod tests {
+    use super::*;
+
+    #[test]
+    fn test_6() {
+        assert_eq!(String::from("PAHNAPLSIIGYIR"), Solution::convert(
+            String::from("PAYPALISHIRING"), 3
+        ));
+    }
+}