c++合并动态数组vector，求中位数

Given two sorted arrays nums1 and nums2 of size m and n respectively, return the median of the two sorted arrays.

The overall run time complexity should be O(log (m+n)).

Example 1:

Input: nums1 = [1,3], nums2 = [2]
Output: 2.00000
Explanation: merged array = [1,2,3] and median is 2.
Example 2:

Input: nums1 = [1,2], nums2 = [3,4]
Output: 2.50000
Explanation: merged array = [1,2,3,4] and median is (2 + 3) / 2 = 2.5.

Constraints:

nums1.length == m
nums2.length == n
0 <= m <= 1000
0 <= n <= 1000
1 <= m + n <= 2000
-106 <= nums1[i], nums2[i] <= 106

class Solution {
public:
    double findMedianSortedArrays(vector& nums1, vector& nums2) {
        int i = 0, j = 0;
        int idx1, idx2, m = (int)nums1.size(), n = (int)nums2.size(), total =m+n;
        //合并
        vector res;
        while((i != m) && (j != n)) {
            if(nums1[i] < nums2[j]) res.push_back(nums1[i++]);
            else res.push_back(nums2[j++]);
        }
        while(i != m) res.push_back(nums1[i++]);
        while(j != n) res.push_back(nums2[j++]);
		//排序
        if(total % 2 == 1) idx1 = idx2 = total / 2;
        else {
            idx1 = (total / 2) - 1;
            idx2 = idx1 + 1;
        }
        float median = (res[idx1] + res[idx2]) / 2.0;
        return median;
    }
};