每天一道算法题--Median of Two Sorted Arrays

最新推荐文章于 2022-09-18 18:35:40 发布

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本文介绍了一种在O(log(m+n))的时间复杂度下找到两个已排序数组中位数的方法。通过对比每个位置的元素，将两个数组合并为一个有序数组，然后根据数组长度的奇偶性计算中位数。

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)).

You may assume nums1 and nums2 cannot be both empty.

Example 1:

nums1 = [1, 3]
nums2 = [2]

The median is 2.0

Example 2:

nums1 = [1, 2]
nums2 = [3, 4]

The median is (2 + 3)/2 = 2.5

解题思路：抓住关键：sorted。已经排序好了，所以只需对每个位置比较判断就好了。

代码：

double findMedianSortedArrays(vector<int>& nums1, vector<int>& nums2)
{
    vector<int> tmp;
    int len1 =nums1.size(),len2 = nums2.size();
    //cout<<len1<<" "<<len2<<endl;
    int i =0,j=0;
    while(i<len1&&j<len2){
        if(nums1[i]<nums2[j]){
            tmp.push_back(nums1[i]);
            i++;
        }else if(nums1[i]>nums2[j]){
            tmp.push_back(nums2[j]);
            j++;
        }else{
            tmp.push_back(nums1[i]);
            tmp.push_back(nums2[j]);
            i++;
            j++;
        }
    }
    //cout<<tmp.size()<<endl;
    while(i<len1){
        tmp.push_back(nums1[i++]);
    }
    while(j<len2){
        tmp.push_back(nums2[j++]);
    }
    int length = tmp.size();
    //cout<<length<<endl;
    if(length%2!=0)
        return 1.0*tmp[length/2];
    else
        return (tmp[length/2]+tmp[length/2-1])/2.0;
}