LeetCode26:Median of Two Sorted Arrays
2015-01-31 16:28
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【题目】
There are two sorted arrays A and B 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)).
【分析】
这个题目按照一般的思路,是先将两个数组归并,然后找到中位数。时间复杂度大概是O(N logM).因此提供一个新的思路:如果A[k/2-1]<B[k/2-1],那么A[0]~A[k/2-1]一定在第k小的数的序列当中,可以用反证法证明。
这种思路可以推广到求合并后的第k大数。在这里k为(m+n)/2。
class Solution {
public:
double findKth(int A[], int m, int B[], int n, int k)
{
//m is equal or smaller than n
if (m > n)
return findKth(B, n, A, m, k);
if (m == 0)
return B[k-1];
if (k <= 1)
return min(A[0], B[0]);
int pa = min(k / 2, m), pb = k - pa;
if (A[pa-1] < B[pb-1])
{
return findKth(A + pa, m - pa, B, n, k - pa);
}
else if(A[pa-1] > B[pb-1])
{
return findKth(A, m, B + pb, n - pb, k - pb);
} else
return A[pa-1];
}
double findMedianSortedArrays(int A[], int m, int B[], int n) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
int k = m + n;
if (k & 0x1)
{
return findKth(A, m, B, n, k / 2 + 1);
} else
{
return (findKth(A, m, B, n, k / 2) + findKth(A, m, B, n, k / 2 + 1)) / 2;
}
}
};
There are two sorted arrays A and B 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)).
【分析】
这个题目按照一般的思路,是先将两个数组归并,然后找到中位数。时间复杂度大概是O(N logM).因此提供一个新的思路:如果A[k/2-1]<B[k/2-1],那么A[0]~A[k/2-1]一定在第k小的数的序列当中,可以用反证法证明。
这种思路可以推广到求合并后的第k大数。在这里k为(m+n)/2。
class Solution {
public:
double findKth(int A[], int m, int B[], int n, int k)
{
//m is equal or smaller than n
if (m > n)
return findKth(B, n, A, m, k);
if (m == 0)
return B[k-1];
if (k <= 1)
return min(A[0], B[0]);
int pa = min(k / 2, m), pb = k - pa;
if (A[pa-1] < B[pb-1])
{
return findKth(A + pa, m - pa, B, n, k - pa);
}
else if(A[pa-1] > B[pb-1])
{
return findKth(A, m, B + pb, n - pb, k - pb);
} else
return A[pa-1];
}
double findMedianSortedArrays(int A[], int m, int B[], int n) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
int k = m + n;
if (k & 0x1)
{
return findKth(A, m, B, n, k / 2 + 1);
} else
{
return (findKth(A, m, B, n, k / 2) + findKth(A, m, B, n, k / 2 + 1)) / 2;
}
}
};
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