diff --git a/src/solution/mod.rs b/src/solution/mod.rs
index bf37cc0..8e9a951 100644
--- a/src/solution/mod.rs
+++ b/src/solution/mod.rs
@@ -4,4 +4,5 @@ mod s0020_valid_parentheses;
 mod s2697_lexicographically_smallest_palindrome;
 mod s0002_add_two_numbers;
 mod s0003_longest_substring_without_repeating_characters;
+mod s0162_find_peak_element;
 mod s2828_check_if_a_string_is_an_acronym_of_words;
diff --git a/src/solution/s0162_find_peak_element.rs b/src/solution/s0162_find_peak_element.rs
new file mode 100644
index 0000000..046f995
--- /dev/null
+++ b/src/solution/s0162_find_peak_element.rs
@@ -0,0 +1,82 @@
+/**
+ * [162] Find Peak Element
+ *
+ * A peak element is an element that is strictly greater than its neighbors.
+ * Given a 0-indexed integer array nums, find a peak element, and return its index. If the array contains multiple peaks, return the index to any of the peaks.
+ * You may imagine that nums[-1] = nums[n] = -∞. In other words, an element is always considered to be strictly greater than a neighbor that is outside the array.
+ * You must write an algorithm that runs in O(log n) time.
+ *  
+ * <strong class="example">Example 1:
+ * 
+ * Input: nums = [1,2,3,1]
+ * Output: 2
+ * Explanation: 3 is a peak element and your function should return the index number 2.
+ * <strong class="example">Example 2:
+ * 
+ * Input: nums = [1,2,1,3,5,6,4]
+ * Output: 5
+ * Explanation: Your function can return either index number 1 where the peak element is 2, or index number 5 where the peak element is 6.
+ *  
+ * Constraints:
+ * 
+ * 	1 <= nums.length <= 1000
+ * 	-2^31 <= nums[i] <= 2^31 - 1
+ * 	nums[i] != nums[i + 1] for all valid i.
+ * 
+ */
+pub struct Solution {}
+
+// problem: https://leetcode.cn/problems/find-peak-element/
+// discuss: https://leetcode.cn/problems/find-peak-element/discuss/?currentPage=1&orderBy=most_votes&query=
+
+// submission codes start here
+
+impl Solution {
+    pub fn find_peak_element(nums: Vec<i32>) -> i32 {
+        if nums.len() == 1 {
+            0
+        }
+
+        let (mut left, mut right) = (0, nums.len() - 1);
+
+        let compare = |x| {
+            if x == 0 {
+                nums[0] > nums[1]
+            } else if x == nums.len() - 1 {
+                nums[x] > nums[x - 1]
+            } else {
+                nums[x - 1] < nums[x] && nums[x] > nums[x + 1]
+            }
+        };
+
+        while left <= right {
+            let mid = (left + right) / 2;
+
+            if compare(mid) {
+                return mid as i32
+            }
+
+            if nums[mid] < nums[mid + 1] {
+                left += 1;
+            } else {
+                right -= 1;
+            }
+        }
+
+        -1
+    }
+}
+
+// submission codes end
+
+#[cfg(test)]
+mod tests {
+    use super::*;
+
+    #[test]
+    fn test_162() {
+        assert_eq!(2, Solution::find_peak_element(vec![1,2,3,1]));
+        assert_eq!(5, Solution::find_peak_element(vec![1,2,1,3,5,6,4]));
+        assert_eq!(1, Solution::find_peak_element(vec![1,2]));
+    }
+}