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)).

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class Solution {
public:
    /*
     * 首先的想法是用额外的一个vector保存排好序的数组,快速排序 o(nlogn)
     *
     */
    double _findMedianSortedArrays(vector<int>& nums1, vector<int>& nums2) {
        vector<int>::iterator it;
        vector<int> all_nums;

        for (it=nums1.begin(); it!=nums1.end(); it++) {
            all_nums.push_back(*it);
        }

        for (it=nums2.begin(); it!=nums2.end(); it++) {
            all_nums.push_back(*it);
        }

        sort(all_nums.begin(), all_nums.end());

        int len_all_nums = all_nums.size();
        if (len_all_nums % 2 == 1) {
            return all_nums[len_all_nums / 2];
        } else {
            double mid_left = all_nums[len_all_nums / 2 - 1];
            double mid_right = all_nums[len_all_nums / 2];
            double mid_num = (mid_left + mid_right) / 2;
            return mid_num;
        }
    }
    /*
     * 也可以转换为找到两个排序链表的第i个数, 不过太麻烦了
     * 1.
     * */
   double findMedianSortedArrays(vector<int>& nums1, vector<int>& nums2) {
        int all_len = nums1.size() + nums2.size();
        int first_index = -1;
        int second_index = -1;
        if (all_len % 2 == 1) {
            first_index = all_len / 2;
        } else {
            first_index = all_len / 2 - 1;
            second_index = all_len / 2;
        }
        double first_num = 0;
        double second_num = 0;
        //下面就是找到对应的数
        int n1 = 0;
        int n2 = 0;
        int find_index = 0;
        for (;find_index<first_index; find_index++) {
            if ((n1 < nums1.size()) && (n2< nums2.size())) {
                if (nums1[n1] <= nums2[n2]) {
                    n1++;
                } else {
                    n2++;
                }
            } else if(n1 < nums1.size()) {
                n1++;
            } else {
                n2++;
            }
        }

        if (n1 == nums1.size()) {
            first_num = nums2[n2];
            if (second_index != -1) {
                second_num = nums2[n2 + 1];
            }
        } else if (n2 == nums2.size()) {
            first_num = nums1[n1];
            if (second_index != -1) {
                second_num = nums1[n1 + 1];
            }
        } else {
            /*
            first_num = min(nums1[n1], nums2[n2]);
            //second_num = max(nums1[n1], nums2[n2]);
            if ((n1 + 1) < nums1.size()) {
                second_num = min(nums1[n1 + 1], nums2[n2]);
            } else {
                second_num = nums2[n2];
            }
            */
            if (nums1[n1] <= nums2[n2]) {
                first_num = nums1[n1];
                if ((n1 + 1) < nums1.size()) {
                    second_num = min(nums1[n1 + 1], nums2[n2]);
                } else {
                    second_num = nums2[n2];
                }
            } else {
                first_num = nums2[n2];
                if ((n2 + 1) < nums2.size()) {
                    second_num = min(nums1[n1], nums2[n2 + 1]);
                } else {
                    second_num = nums1[n1];
                }
            }
        }

        if (second_index != -1) {
            return (first_num + second_num) / 2.0;
        } else {
            return first_num;
        }
   }
};