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 2.0
Example 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;
}