4. Median of Two Sorted Arrays

最新推荐文章于 2024-04-05 16:34:21 发布

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本文介绍了一个高效算法，用于找到两个已排序数组的中位数，并确保总运行时间复杂度为O(log(m+n))。通过示例说明了算法的工作原理及实现过程。

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)).

Example 1:

nums1 = [1, 3]
nums2 = [2]
The median is 2.0

Example 2:

nums1 = [1, 4]
nums2 = [3, 4]
The median is (2+3)/2=2.5

思路：这道题目是限制了时间复杂度的，想了很久没做出来，便直接在网上看参考了，https://discuss.leetcode.com/topic/16797/very-concise-o-log-min-m-n-iterative-solution-with-detailed-explanation/2，代码如下：

class Solution {
public:
    double findMedianSortedArrays(vector<int>& nums1, vector<int>& nums2) {
        int m = nums1.size(), n = nums2.size();
        if (m < n) return findMedianSortedArrays(nums2, nums1);
        if (n == 0) return ((double)nums1[(m - 1) / 2] + (double)nums1[m / 2]) / 2.0;
        int left = 0, right = n * 2;
        while (left <= right) {
            int mid2 = (left + right) / 2;
            int mid1 = m + n - mid2;
            double L1 = mid1 == 0 ? INT_MIN : nums1[(mid1 - 1) / 2];
            double L2 = mid2 == 0 ? INT_MIN : nums2[(mid2 - 1) / 2];
            double R1 = mid1 == m * 2 ? INT_MAX : nums1[mid1 / 2];
            double R2 = mid2 == n * 2 ? INT_MAX : nums2[mid2 / 2];
            if (L1 > R2) left = mid2 + 1;
            else if (L2 > R1) right = mid2 - 1;
            else return (max(L1, L2) + min(R1, R2)) / 2;
        }
        return -1;
    }
};