Leetcode 4 Median of Two Sorted Arrays

 1 class Solution {
 2 public:
 3     double findMedianSortedArrays(int A[], int m, int B[], int n) {
 4         int middle ;
 5         int temp = (m+n)%2;
 6         if(temp == 1)
 7             middle = (m+n)/2;
 8         else
 9             middle=(m+n)/2-1;
10         int i=0,j=0;
11         int count=0;
12         int flag=0;
13         double res;
14         while (count<middle&&i<m&&j<n)
15         {
16             if(A[i]<B[j]){
17                 i++;
18             }
19             else if (A[i]>B[j])
20             {
21                 j++;
22             }else
23             //当A[i]==B[j]时，两个数组下标移动需要取决于下一个数，比如{1,1}和{1,2}；
24             //1==1因此需要i++，如果是{1,2}和{1,1}，则j++
25             {
26                 if(i+1<m&&A[i+1]==A[i])
27                     i++;
28                 else if(j+1<n&&B[j+1]==B[j])
29                     j++;
30                 else
31                     i++;
32 
33 
34             }
35             count++;
36         }
37         
38         if (i==m)//数组A已经越界
39         {
40             if(temp==0){
41                 res = (B[j+middle-count]+B[j+middle-count+1])/2.0;
42             }else
43                 res = B[j+middle-count];
44 
45         }else if (j==n)//数组B已经越界
46         {
47             if(temp==0){
48                 res = (A[i+middle-count]+A[i+middle-count+1])/2.0;
49             }else
50                 res = A[i+middle-count];
51         }else
52         {
53             if (temp == 0)
54             {
55                 if(i+1<m && A[i+1]<B[j])
56                     res = (A[i]+A[i+1])/2.0;
57                 else if(j+1<n && B[j+1]<A[i])
58                     res = (B[j]+B[j+1])/2.0;
59                 else
60                     res = (B[j]+A[i])/2.0;
61 
62             }else
63             {
64                 res = A[i]>B[j] ? B[j]:A[i];
65             }
66         }
67 
68         return  res;
69 
70     }
71 };