diff --git a/C++/004.Median of Two Sorted Arrays.cpp b/C++/004.Median of Two Sorted Arrays.cpp
new file mode 100644
index 0000000..c1979e7
--- /dev/null
+++ b/C++/004.Median of Two Sorted Arrays.cpp	
@@ -0,0 +1,52 @@
+4. Median of Two Sorted Arrays
+There are two sorted arrays nums1 and nums2 of size m and n respectively.
+
+Find the median of the two sorted arrays. The overall run time complexity should be O(log (m+n)).
+
+You may assume nums1 and nums2 cannot be both empty.
+
+Example 1:
+
+nums1 = [1, 3]
+nums2 = [2]
+
+The median is 2.0
+Example 2:
+
+nums1 = [1, 2]
+nums2 = [3, 4]
+
+The median is (2 + 3)/2 = 2.5
+
+
+class Solution {
+public:
+    double findMedianSortedArrays(vector<int>& nums1, vector<int>& nums2) {
+        int len1 = nums1.size();
+        int len2 = nums2.size();
+        int mid_len = (int)(len1+len2) /2+1;
+        //create an array to store the values of half of nums1 and nums2.
+        float store[mid_len] = {0};
+        int k =0;
+        int i=0,j=0;
+        while(i<len1&&j<len2&&k<mid_len){
+            if(nums1[i]<=nums2[j]){
+                store[k++] = nums1[i++];
+            }
+            else if(nums2[j]<=nums1[i]){
+                store[k++] = nums2[j++];
+            }
+        }
+        while(i<len1&&k<mid_len){
+            store[k++] = nums1[i++];
+
+        }
+         while(j<len2&&k<mid_len){
+                store[k++] = nums2[j++];
+        }
+        if((len1+len2)%2==1)
+            return store[mid_len-1];
+        else
+            return (store[mid_len-1]+store[mid_len-2]) /2;
+    }
+};
diff --git a/C++/008. String to Integer (atoi).cpp b/C++/008. String to Integer (atoi).cpp
index b552567..038cee7 100644
--- a/C++/008. String to Integer (atoi).cpp	
+++ b/C++/008. String to Integer (atoi).cpp	
@@ -1,4 +1,4 @@
-8. String to Integer (atoi)   
+8. String to Integer (atoi)
 My Submissions QuestionEditorial Solution
 Total Accepted: 99420 Total Submissions: 737262 Difficulty: Easy
 Implement atoi to convert a string to an integer.
@@ -46,6 +46,13 @@ class Solution {
                 break;
             }
         }
+        //remove the '0' in front,wich is useless, to reduce the length
+        // the input “0000000002000” will not pass now if without the operation
+        int j=0;
+        while(str[j]=='0')
+            j++;
+        str = str.substr(j);
+
         if(str.length() == 0)
             return 0;
         if(str.length() > 10) {
@@ -65,4 +72,4 @@ class Solution {
             return (int)ans;
         }
     }
-};
\ No newline at end of file
+};