LeetCode Java First 400 题解-004_leetcode 004 java-优快云博客

本文链接：https://blog.youkuaiyun.com/felomeng/article/details/102699200

本文介绍了一种在O(log(m+n))的时间复杂度下找到两个有序数组中位数的方法。通过将问题转化为寻找第k个元素，利用递归思想在两个数组中逐步缩小搜索范围，最终找到中位数。

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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)).

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

public double findMedianSortedArrays(int[] nums1, int[] nums2) {

    int m = nums1.length, n = nums2.length;

    int l = (m + n + 1) >> 1;

    int r = (m + n + 2) >> 1;

    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) return B[bStart + k - 1];

    if (bStart == B.length) 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);

}

思路：转化为求第k个数，最中间两数的平均即是要求值，

结束条件：A或B结束，从另一个数组中取第k个数；或，k=1，取两者中较小的数。

推理条件：如果单个数组k/2位存在，则移k/2（记为临时中位数），临时中位数较小的向前移k/2