LeetCode 4. Median of Two Sorted Arrays [Hard]

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题目来源：

题目难度：Hard

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:

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nums1 = [1, 3]
nums2 = [2]

The median is 2.0

Example 2:

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nums1 = [1, 2]
nums2 = [3, 4]

The median is (2 + 3)/2 = 2.5

解答1[Java]

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public double findMedianSortedArrays(int[] nums1, int[] nums2) {
    int[] nums = new int[nums1.length + nums2.length];
    System.arraycopy(nums1, 0, nums, 0, nums1.length);
    System.arraycopy(nums2, 0, nums, nums1.length, nums2.length);
    Arrays.sort(nums);
    if ((nums.length & 1) == 0) {
        return (nums[(nums.length / 2) - 1] + nums[nums.length / 2]) / 2.0;
    } else {
        return nums[nums.length / 2];
    }
}

时空复杂度分析

时间复杂度：$O((m+n)log(m+n))$

空间复杂度：$O(m+n)$

解法2[Java]

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class Solution {
    public double findMedianSortedArrays(int[] A, int[] B) {
        int m = A.length;
        int n = B.length;
        if (m > n) {
            int[] temp = A;
            A = B;
            B = temp;
            int tmp = m;
            m = n;
            n = tmp;
        }
        int iMin = 0, iMax = m, halfLen = (m + n + 1) / 2;
        while (iMin <= iMax) {
            int i = (iMin + iMax) / 2;
            int j = halfLen - i;
            if (i < iMax && B[j - 1] > A[i]) {
                iMin = i + 1;
            } else if (i > iMin && A[i - 1] > B[j]) {
                iMax = i - 1;
            } else {
                int maxLeft = 0;
                if (i == 0) {
                    maxLeft = B[halfLen - 1];
                } else if (j == 0) {
                    maxLeft = A[halfLen - 1];
                } else {
                    maxLeft = Math.max(A[i - 1], B[j - 1]);
                }
                if (((m + n) & 1) == 1) {
                    return maxLeft;
                }

                int minRight;
                if (i == m) {
                    minRight = B[j];
                } else if (j == n) {
                    minRight = A[i];
                } else {
                    minRight = Math.min(B[j], A[i]);
                }

                return (maxLeft + minRight) / 2.0;
            }
        }
        return 0.0;
    }
}

算法分析

核心思想是二分查找。

假设 A 数组是长度为 4，B 数组长度为 6，那么把 A 数组和 B 数组混合，然后重新排序，称之为数组 C。那么数组 C 的前半部分，必定由 A 中的若干个数和 B 中的若干个数组成。

时空复杂度分析

时间复杂度：$O(log(min(m,n))$

空间复杂度：$O(1)$

拓展

思考：如果是求三个有序数组的中位数呢？

参考资料

  1. 官方题解：Median of Two Sorted Arrays - LeetCode Articles