leetcode题解-4. Median of Two Sorted Arrays

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 2.0Example 2:nums1 = [1, 2]nums2 = [3, 4]The median is (2 + 3)/2 = 2.5

题目就是要寻找两个排序数组的中位数、，要分为两种情况考虑，一种是共有奇数个数，那么中位数就是中间那个，否则是中间两个数的平均值。所以我们也分为两种情况进行编程实现。

public double findMedianSortedArrays(int[] A, int[] B) {        int m = A.length, n = B.length;        int l = (m + n + 1) / 2;        int r = (m + n + 2) / 2;        return (getkth(A, 0, B, 0, l) + getkth(A, 0, B, 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);// Check: aRight + bLeft     else         return getkth(A, aStart,       B, bStart + k/2, k - k/2);// Check: bRight + aLeft}

另外一种思路就是使用二分法，然后在进行判断要返回的值：

    public double findMedianSortedArrays(int A[], int B[]) {        int n = A.length;        int m = B.length;        if (n > m)            return findMedianSortedArrays(B, A);        // now, do binary search        int k = (n + m - 1) / 2;        int l = 0, r = Math.min(k, n); // r is n, NOT n-1, this is important!!        while (l < r) {            int midA = (l + r) / 2;            int midB = k - midA;            if (A[midA] < B[midB])                l = midA + 1;            else                r = midA;        }        // after binary search, we almost get the median because it must be between        // these 4 numbers: A[l-1], A[l], B[k-l], and B[k-l+1]        // if (n+m) is odd, the median is the larger one between A[l-1] and B[k-l].        // and there are some corner cases we need to take care of.        int a = Math.max(l > 0 ? A[l - 1] : Integer.MIN_VALUE, k - l >= 0 ? B[k - l] : Integer.MIN_VALUE);        if (((n + m) & 1) == 1)            return (double) a;        // if (n+m) is even, the median can be calculated by        //      median = (max(A[l-1], B[k-l]) + min(A[l], B[k-l+1]) / 2.0        // also, there are some corner cases to take care of.        int b = Math.min(l < n ? A[l] : Integer.MAX_VALUE, k - l + 1 < m ? B[k - l + 1] : Integer.MAX_VALUE);        return (a + b) / 2.0;    }