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Median of Two Sorted Arrays

2015-06-22 00:00 423 查看
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)).

public class Solution {

public  double findMedianSortedArrays(int A[], int B[]) {
int m = A.length;
int n = B.length;

if ((m + n) % 2 != 0) // odd
return (double) findKth(A, B, (m + n) / 2, 0, m - 1, 0, n - 1);
else { // even
return (findKth(A, B, (m + n) / 2, 0, m - 1, 0, n - 1)
+ findKth(A, B, (m + n) / 2 - 1, 0, m - 1, 0, n - 1)) * 0.5;
}
}

public int findKth(int A[], int B[], int k,
int aStart, int aEnd, int bStart, int bEnd) {

int aLen = aEnd - aStart + 1;
int bLen = bEnd - bStart + 1;

// Handle special cases
if (aLen == 0)
return B[bStart + k];
if (bLen == 0)
return A[aStart + k];
if (k == 0)
return A[aStart] < B[bStart] ? A[aStart] : B[bStart];

int aMid = aLen * k / (aLen + bLen); // a's middle count
int bMid = k - aMid - 1; // b's middle count

// make aMid and bMid to be array index
aMid = aMid + aStart;
bMid = bMid + bStart;

if (A[aMid] > B[bMid]) {
k = k - (bMid - bStart + 1);
aEnd = aMid;
bStart = bMid + 1;
} else {
k = k - (aMid - aStart + 1);
bEnd = bMid;
aStart = aMid + 1;
}

return findKth(A, B, k, aStart, aEnd, bStart, bEnd);
}
}
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标签:  java leetcode