【算法day3】寻找两个正序数组的中位数

寻找两个正序数组的中位数

给定两个大小分别为 m 和 n 的正序（从小到大）数组 nums1 和 nums2。请你找出并返回这两个正序数组的 中位数 。

算法的时间复杂度应该为 O(log (m+n)) 。

https://leetcode.cn/problems/median-of-two-sorted-arrays/description/

class Solution {
public:
    double findMedianSortedArrays(vector<int>& nums1, vector<int>& nums2) {
        int mid = (nums1.size() + nums2.size()) / 2 - 1;
        int odds = (nums1.size() + nums2.size()) % 2;
        int mid_num1 = 0, mid_num2 = 0;

        int i = 0, j = 0, k = 0;
        // 偶数取均值，奇数取中间那个值
        while (i < nums1.size() && j < nums2.size() && k <= mid) {
            if (nums1[i] < nums2[j]) {
                mid_num1 = nums1[i];
                i++;
            } else {
                mid_num1 = nums2[j];
                j++;
            }
            k++;
        }
        while (i < nums1.size() && k <= mid) {
            mid_num1 = nums1[i];
            i++;
            k++;
        }
        while (j < nums2.size() && k <= mid) {
            mid_num1 = nums2[j];
            j++;
            k++;
        }
        if (i < nums1.size() && j < nums2.size()) {
            if (nums1[i] < nums2[j]) {
                mid_num2 = nums1[i];
            } else {
                mid_num2 = nums2[j];
            }
        } else if (i < nums1.size()) {
            mid_num2 = nums1[i];
        } else if (j < nums2.size()) {
            mid_num2 = nums2[j];
        }
        if (odds % 2 == 0) {
            float result = mid_num1 + mid_num2;
            return result / 2;
        } else {
            return mid_num2;
        }
    }
};