leetcode:Median of Two Sorted Arrays

最新推荐文章于 2023-06-12 16:30:30 发布

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本文介绍了一个算法，用于找到两个已排序数组的中位数，其时间复杂度为 O(log(m+n))。该算法通过递归地比较数组的中间元素来实现。

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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 {
    
private:
    double findKthSortedArrays(int nums1[], int m, int nums2[], int n, int k) {
        
        if (m > n)
            return findKthSortedArrays(nums2, n, nums1, m, k);
            
        if (m == 0)
            return nums2[k-1];
            
        if (k == 1)
            return min(nums1[0], nums2[0]);
            
        int p1 = min(k/2, m);
        int p2 = k-p1;
        
        if (nums1[p1-1] == nums2[p2-1])
            return nums1[p1-1];
        else if(nums1[p1-1] < nums2[p2-1])
            return findKthSortedArrays(nums1+p1, m-p1, nums2, n, k-p1);
        else
            return findKthSortedArrays(nums1, m, nums2+p2, n-p2, k-p2);
        
    }
    
public:
    double findMedianSortedArrays(vector<int>& nums1, vector<int>& nums2) {
        
        int totalLen = nums1.size() + nums2.size();
        int median;
        if (totalLen % 2)
        {
            return findKthSortedArrays(&nums1[0], nums1.size(), &nums2[0], nums2.size(), totalLen/2+1); 
        }
        else
        {
            return (findKthSortedArrays(&nums1[0], nums1.size(), &nums2[0], nums2.size(), totalLen/2) + findKthSortedArrays(&nums1[0], nums1.size(), &nums2[0], nums2.size(), totalLen/2+1)) / 2;
        }
    }
};