Median of Two Sorted Arrays

最新推荐文章于 2023-10-07 04:23:19 发布

原创最新推荐文章于 2023-10-07 04:23:19 发布 · 383 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#algorithm

算法专栏收录该内容

1 篇文章

订阅专栏

本文介绍了一种高效算法，用于查找两个已排序数组中的中位数，并将时间复杂度控制在O(log(m+n))范围内。该算法还进一步扩展到寻找第K小的元素。通过递归划分和比较两个数组的中间元素来缩小搜索范围。

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)).

延伸为求第K小的数

if (aMid < bMid) Keep [aRight + bLeft]

else Keep [bRight + aLeft]

public class _4_MedianOfTwoSortedArrays {
    public double findMedianSortedArrays(int[] nums1, int[] nums2) {
        int m = nums1.length;
        int n = nums2.length;
        int l = (m + n + 1) / 2;
        int r = (m + n + 2) / 2;
        return (getKth(nums1, 0, nums2, 0, l) + getKth(nums1, 0, nums2, 0, r)) / 2.0;
    }

    public double getKth(int[] A, int aStart, int[] B, int bStart, int k) {
        if (aStart > A.length - 1) return B[bStart + k - 1];
        if (bStart > B.length - 1) return A[aStart + k - 1];
        if (k == 1) return Math.min(A[aStart], B[bStart]);

        int aMid = Integer.MAX_VALUE, bMid = Integer.MAX_VALUE;
        if (aStart + k / 2 - 1 < A.length) aMid = A[aStart + k / 2 - 1];
        if (bStart + k / 2 - 1 < B.length) bMid = B[bStart + k / 2 - 1];

        if (aMid < bMid)
            return getKth(A, aStart + k / 2, B, bStart, k - k / 2);
        else
            return getKth(A, aStart, B, bStart + k / 2, k - k / 2);
    }

//    public double getKth(int[] A, int aStart, int[] B, int bStart, int k) {
//        if (aStart > A.length - 1) return B[bStart + k - 1];
//        if (bStart > B.length - 1) return A[aStart + k - 1];
//        if (k == 1) return Math.min(A[aStart], B[bStart]);
//
//        int aMid = Integer.MAX_VALUE;
//        int bMid = Integer.MAX_VALUE;
//        if (aStart + k / 2 - 1 < A.length) aMid = A[aStart + k / 2 - 1];
//        if (bStart + k / 2 - 1 < B.length) bMid = B[bStart + k / 2 - 1];
//
//        if (aMid < bMid) return getKth(A, aStart + k / 2, B, bStart, k - k / 2);
//        else return getKth(A, aStart, B, bStart + k / 2, k - k / 2);
//
//    }

    public static void main(String[] args) {
        _4_MedianOfTwoSortedArrays obj = new _4_MedianOfTwoSortedArrays();
        int[] A = {1, 2, 5, 8, 9};
        int[] B = {2, 4, 6, 8, 18};

        System.out.println(obj.getKth(A, 0, B, 0, 6));
    }
}