4. 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)) 。

 

题解：采用了遍历的手法，刚好符合log(m + n)的时间复杂度，要注意两个数组总数和为奇数或者偶数的情况

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#include "../include/leetcode.h"   class Solution { public:     double findMedianSortedArrays(vector<int>& nums1, vector<int>& nums2) {           int left = (nums1.size() + nums2.size()) % 2;           int medianPositioin = (nums1.size() + nums2.size()) / 2;           //两个数组数量加起来是偶数的情况           if (left == 0) {                medianPositioin -= 1;           }             int index = 0;           int i1, i2;           double currentNum;           for (i1 = i2 = 0; i1 < nums1.size() && i2 < nums2.size() && index <= medianPositioin;) {                currentNum = nums1[i1] < nums2[i2] ? nums1[i1] : nums2[i2];                if (currentNum == nums1[i1]) {                     ++i1;                }                else {                     ++i2;                }                index += 1;           }             for (; i1 < nums1.size() && index <= medianPositioin; ++i1) {                currentNum = nums1[i1];                index += 1;           }              for (; i2 < nums2.size() && index <= medianPositioin; ++i2) {                currentNum = nums2[i2];                index += 1;           }             if(left == 1) {                return currentNum;           }           else {                double currentNextNum;                if (i1 < nums1.size() && i2 < nums2.size()) {                     currentNextNum = nums1[i1] < nums2[i2] ? nums1[i1] : nums2[i2];                }                else if (i1 < nums1.size()) {                     currentNextNum = nums1[i1];                }                else if (i2 < nums2.size()){                     currentNextNum = nums2[i2];                }                                 return (currentNextNum + currentNum) / 2;           }           } };   int main() {      Solution s;      vector<int> nums1{};      vector<int> nums2{1};      cout << s.findMedianSortedArrays(nums1, nums2) << endl;      return 0; }