median of two sorted array

leetcode4.c

/*
median of two sorted array
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:
    int Max(int a, int b)
    {
        return a>b?a:b;
    }
    int Min(int a,int b)
    {
        return a<b?a:b;
    }
    double findMedianSortedArrays(vector<int>& nums1, vector<int>& nums2) {
        int m = nums1.size();
        int n = nums2.size();
        if(m>n){
            vector<int> temp;
            temp = nums1;
            nums1 = nums2;
            nums2 = temp;
            int temp2 = m; 
            m = n; 
            n = temp2;
        }
        if(m==0 and n==1){
            return nums2.at(0);
        }
        if(m==1 and n==1){
            return float(nums1.at(0)+nums2.at(0))/2;
        }
        int min = 0, max = m, half = (m+n+1)/2;
        int middle;
        while(min<=max){
            int i = (min+max)/2; //i,j : the length of left A, B;
            int j = half - i;
            if(i < max and nums2.at(j-1) > nums1.at(i)) 
            {
                min = i+1;
            }
            else if (i>min and nums2.at(j) < nums1.at(i-1))
            {
                max = i-1;
            }
            else // find index
            {
                int maxleft;
                //理解i的意思，表示A的左边部分有0个，所以左边最大一定是B中
                //虽然n>m, 但是j也存在边界情况，当两个数组长度相等时，需要判断j==0或者n
                if(i==0){ maxleft = nums2.at(j-1);} // 这里A[i]一定比B[j-1]大，所以A[i]应该在右半部分，如果A[i]< B[j-1] 会满足条件1，使i==1
                else if (j==0) maxleft=nums1.at(i-1);
                else{
                    maxleft = Max(nums2.at(j-1), nums1.at(i-1));
                }
                if ((m+n)%2 == 1) return maxleft;
                int minright;
                if(i==m){minright = nums2.at(j);}
                else if (j==n) minright = nums1.at(i);
                else minright = Min(nums2.at(j), nums1.at(i));
                return double(minright+maxleft)/2;
            }
        }
        return 0;
    }
};