LeetCode | 4. Median of Two Sorted Arrays

最新推荐文章于 2023-06-12 16:30:30 发布

1LOVESJohnny

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问题，即找到两个已排序数组的中位数，整体运行复杂度为O(log(m+n))。通过合并两个数组并查找中位数的方法，展示了如何解决这一问题。例子包括了不同长度的数组情况。

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题目：

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

代码：

class Solution {
public:
    double findMedianSortedArrays(vector<int>& nums1, vector<int>& nums2) {
        int n1 = nums1.size();
		int n2 = nums2.size();
        if( n1 == 0 && n2 == 0)
			return 0;
		vector<int> sum;
		int i1 = 0, i2 = 0;
		while( i1 < n1 && i2 < n2)
		{
			if(nums1[i1] < nums2[i2])
			{
				sum.push_back(nums1[i1]);
				i1++;
			}
			else
			{
				sum.push_back(nums2[i2]);
				i2++;
			}
			if( i1 == n1 && i2 != n2)
			{
				sum.insert(sum.end(), nums2.begin()+i2, nums2.end());
			}
			if( i1 != n1 && i2 == n2)
			{
				sum.insert(sum.end(), nums1.begin()+i1, nums1.end());
			}
		}
        if( n1 == 0 )
			sum = nums2;
		if( n2 == 0 )
			sum = nums1;
		if((n1 + n2) % 2)
		{
			return sum[(n1+n2)/2];
		}
		else
			return (sum[(n1+n2)/2 - 1] + sum[(n1+n2)/2]) / 2.0;
    }
};