題目描述

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)).

題目要求尋找兩個已排序陣列的中位數

解題程式碼

本題我使用了比較傻瓜的方法去解決這個問題。如果兩個陣列均不為空，則將兩個數組合併為一個數組，然後再獲得合併後的陣列的中位數。

class Solution {
public
 
:
    double findMedianSortedArrays(vector<int>& nums1, vector<int>& nums2) {
        if (nums1.empty() && nums2.empty())
        {
            return 0;
        }
        else if (!nums1.empty() && nums2.empty())
        {
            int size = nums1.size();
            if
 
 (size % 2 == 0)
                return (nums1[size / 2] + nums1[size / 2 - 1]) / 2.0;
            else
                return nums1[size / 2];
        }
        else if (nums1.empty() && !nums2.empty())
        {
            int size = nums2.size();
            if (size % 2 == 0)
                return
 
 (nums2[size / 2] + nums2[size / 2 - 1]) / 2.0;
            else
                return nums2[size / 2];
        }

        bool isEven = (nums1.size() + nums2.size()) % 2 == 0;
        int count = (nums1.size() + nums2.size()) / 2 + 1;

        vector<int> union_nums;
        vector<int>::iterator it1 = nums1.begin();
        vector<int>::iterator it2 = nums2.begin();
        vector<int>::iterator it1_end = nums1.end();
        vector<int>::iterator it2_end = nums2.end();
        for(size_t i = 0; i != count; ++i)
        {
            if(it1 != it1_end && it2 != it2_end)
            {
                if(*it1 < * it2)
                {
                    union_nums.push_back(*it1);
                    it1++;
                }
                else
                {
                    union_nums.push_back(*it2);
                    it2++;
                }
            }
            else if (it1 == it1_end)
            {
                union_nums.push_back(*it2);
                it2++;
            }
            else
            {
                union_nums.push_back(*it1);
                it1++;
            }
        }

        if(isEven) return (union_nums[count-1]+union_nums[count-2])/2.0;
        else return union_nums[count-1];
    }
};