Create median_of_two_sorted_array.md

Author: 17020021029

median_of_two_sorted_array.md

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+# Median of Two Sorted Arrays
+希望没人打开这篇，因为这道题我还没A就想睡觉了！！！！最近又有点神经衰弱，emmmm，老是想东想西，不在状态，要好好调整啊
+## Problem
+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)).
+
+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
+## 解题思路
+主要考查如何减少时间复杂度
+## 未A代码
+```
+void quickSort(int* nums,int first,int end){  
+    int temp,l,r;  
+    if(first>=end)return;  
+    temp=nums[first];  
+    l=first;r=end;  
+    while(l<r){  
+        while(l<r && nums[r]>=temp)
+            r--;  
+        if(l<r)
+            nums[l]=nums[r];  
+        while(l<r && nums[l]<=temp)
+            l++;  
+        if(l<r)
+            nums[r]=nums[l];  
+    }  
+    nums[l]=temp;  
+    quickSort(nums,first,l-1);  
+    quickSort(nums,l+1,end);  
+}
+double findMedianSortedArrays(int* nums1, int nums1Size, int* nums2, int nums2Size) {
+    if(nums1Size==0&&nums2Size==0)
+        return 0;
+    else{
+   int new[nums1Size+nums2Size];
+    double result;
+    for(int i=0;i<nums1Size;i++)
+        new[i]=nums1[i];
+    for(int j=0;j<nums2Size;j++)
+    {
+        int count=nums1Size;
+        new[count]=nums2[j];
+        count++;
+    }
+    quickSort(new,0,nums1Size+nums2Size-2);
+    int size=nums1Size+nums2Size;
+    if((nums1Size+nums2Size)%2!=0){
+        int a=size/2-1;
+        int b=size/2;
+      result=((new[a]+new[b])/2);  
+    }
+    else
+        result=(double)new[size/2];
+    return result;
+    }
+}
+```