LeetCode - Algorithms - 4. Median of Two Sorted Arrays

Problem

4. Median of Two Sorted Arrays

Follow up

The overall run time complexity should be O(log (m+n)).

Java

extra space

On the basis of 88. Merge Sorted Array, this is not preferable method, just better than nothing.

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class Solution {
    public double findMedianSortedArrays(int[] nums1, int[] nums2) {
        final int m = nums1.length;
        final int n = nums2.length;
        if (m == 0)
            return (n & 1) == 1 ? nums2[n / 2] : (nums2[n / 2 - 1] + nums2[n / 2]) / 2.0;
        if (n == 0)
            return (m & 1) == 1 ? nums1[m / 2] : (nums1[m / 2 - 1] + nums1[m / 2]) / 2.0;
        int[] aux = merge(nums1, m, nums2, n);
        double median = ((m + n) & 1) == 1 ? aux[(m + n) / 2] : (aux[(m + n) / 2 - 1] + aux[(m + n) / 2]) / 2.0;
        return median;
    }

    private int[] merge(int[] nums1, int m, int[] nums2, int n) {
        int[] aux = new int[m + n];
        if (m == 0) {
            for (int i = 0; i < n; i++)
                aux[i] = nums2[i];
            return aux;
        } else {
            for (int i = 0; i < m; i++)
                aux[i] = nums1[i];
        }
        int idx = m + n - 1;
        m--;
        n--;
        for (; m >= 0 && n >= 0; ) {
            if (nums1[m] < nums2[n]) {
                aux[idx--] = nums2[n];
                n--;
            } else {
                aux[idx--] = nums1[m];
                m--;
            }
        }
        if (n >= 0) {
            for (int i = 0; i <= n; i++)
                aux[i] = nums2[i];
        }
        return aux;
    }
}

Submission Detail

  • 2091 / 2091 test cases passed.
  • Runtime: 2 ms, faster than 99.73% of Java online submissions for Median of Two Sorted Arrays.
  • Memory Usage: 40.1 MB, less than 10.39% of Java online submissions for Median of Two Sorted Arrays.