leetcode 4. Median of Two Sorted Arrays

最新推荐文章于 2025-08-09 17:01:50 发布

景行cmy

最新推荐文章于 2025-08-09 17:01:50 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。通过调整合并策略，确保了算法的效率。示例展示了如何处理不同长度的数组。

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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)).

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

class Solution {
public:
    double findMedianSortedArrays(vector<int>& nums1, vector<int>& nums2)
    {
        int len1 = nums1.size();
        int len2 = nums2.size();
        if(len1>len2) return findMedianSortedArrays(nums2,nums1);
        int k = (len1+len2+1)/2;
        int left = 0;
        int right = len1;
        int mid = 0;
        int m2 = 0;
        while(left<right)
        {
            mid = left + (right-left)/2;
            m2 = k - mid;
            if(nums1[mid]<nums2[m2-1]) left = mid+1;
            else right = mid;
        }
        mid = left;
        m2 = k - mid;
        int c1 = max(mid<=0 ? INT_MIN : nums1[mid-1] , m2<=0 ? INT_MIN :nums2[m2-1] );
        if((len1+len2)%2==1) return c1;
        int c2 = min(mid>=len1 ? INT_MAX : nums1[mid],m2>=len2 ? INT_MAX : nums2[m2]);
        return (c1+c2)/2.0;
    }
};