diff --git a/0004-median-of-two-sorted-arrays/Code/1.java b/0004-median-of-two-sorted-arrays/Code/1.java
index cfb36cfd..1ca634e4 100644
--- a/0004-median-of-two-sorted-arrays/Code/1.java
+++ b/0004-median-of-two-sorted-arrays/Code/1.java
@@ -1,6 +1,6 @@
 public class Solution {
   public double findMedianSortedArrays(int[] nums1, int[] nums2) {
-    // 使nums1成为较短数组,不仅可以提高检索速度,同时可以避免一些边界问题
+    // Ensure nums1 is the smaller array
     if (nums1.length > nums2.length) {
       int[] temp = nums1;
       nums1 = nums2;
@@ -9,46 +9,39 @@ public double findMedianSortedArrays(int[] nums1, int[] nums2) {
 
     int len1 = nums1.length;
     int len2 = nums2.length;
-    int leftLen = (len1 + len2 + 1) / 2; //两数组合并&排序后,左半边的长度
+    int leftLen = (len1 + len2 + 1) / 2; // Half of the total length (rounded up)
     
-    // 对数组1进行二分检索
     int start = 0;
     int end = len1;
     while (start <= end) {
-      // 两个数组的被测数A,B的位置(从1开始计算)
-      // count1 = 2 表示 num1 数组的第2个数字
-      // 比index大1
-      int count1 = start + ((end - start) / 2);
+      // Partition indices in nums1 and nums2
+      int count1 = start + (end - start) / 2;
       int count2 = leftLen - count1;
       
       if (count1 > 0 && nums1[count1 - 1] > nums2[count2]) {
-        // A比B的next还要大
+        // Move partition in nums1 to the left
         end = count1 - 1;
       } else if (count1 < len1 && nums2[count2 - 1] > nums1[count1]) {
-        // B比A的next还要大
+        // Move partition in nums1 to the right
         start = count1 + 1;
       } else {
-        // 获取中位数
-        int result =  (count1 == 0)? nums2[count2 - 1]: // 当num1数组的数都在总数组右边
-                      (count2 == 0)? nums1[count1 - 1]: // 当num2数组的数都在总数组右边
-                      Math.max(nums1[count1 - 1], nums2[count2 - 1]); // 比较A,B
-        if (isOdd(len1 + len2)) {
-          return result;
+        // We have found the correct partition
+        int result = (count1 == 0) ? nums2[count2 - 1] : // If count1 is 0, take from nums2
+                     (count2 == 0) ? nums1[count1 - 1] : // If count2 is 0, take from nums1
+                     Math.max(nums1[count1 - 1], nums2[count2 - 1]); // Maximum of left side
+        if ((len1 + len2) % 2 == 1) {
+          return result; // If odd, return the max of the left side
         }
 
-        // 处理偶数个数的情况
-        int nextValue = (count1 == len1) ? nums2[count2]:
-                        (count2 == len2) ? nums1[count1]:
+        // If even, find the next smallest element from the right side
+        int nextValue = (count1 == len1) ? nums2[count2] : 
+                        (count2 == len2) ? nums1[count1] : 
                         Math.min(nums1[count1], nums2[count2]);
-        return (result + nextValue) / 2.0;
+
+        return (result + nextValue) / 2.0; // Return the average of left and right sides
       }
     }
 
-    return Integer.MIN_VALUE; // 绝对到不了这里
-  }
-
-  // 奇数返回true,偶数返回false
-  private boolean isOdd(int x) {
-    return (x & 1) == 1;
+    return Integer.MIN_VALUE; // This should never happen if input arrays are valid
   }
 }