log(m+n)找第k大
递归
int find_kth(vector<int>& nums1, int begin1, int size1, vector<int>& nums2, int begin2, int size2, int k)
{
size1 = min(k, size1);//第k大最多只要前k个
size2 = min(k, size2);
if (k == 1)
{
return min(nums1[begin1], nums2[begin2]);
}
if (size1 == 1)
{
if (begin2 + k - 1 < nums2.size())
return min(max(nums1[begin1], nums2[begin2 + k - 2]), nums2[begin2 + k - 1]);
else
return max(nums1[begin1], nums2[begin2 + k - 2]);
}
if (size2 == 1)
{
if (begin1 + k - 1 < nums1.size())
return min(max(nums2[begin2], nums1[begin1 + k - 2]), nums1[begin1 + k - 1]);
else
return max(nums2[begin1], nums1[begin1 + k - 2]);
}
double s = k / static_cast<double>(size1 + size2);//对应的比例位置
int q = s*(size1)+begin1;/**/
int p = s*(size2)+begin2;/**/
if (static_cast<int>(s*(size1)) + static_cast<int>(s*(size2))> k - 1 && (q - begin1) && (p - begin2))
//调节使 k刚好落在p 或 k
{
--p;
--q;
}
if (static_cast<int>(s*(size1)) + static_cast<int>(s*(size2))< k - 3 && (q - begin1) && (p - begin2))
//调节使 k刚好落在p 或 k
{
++p;
++q;
}
if (nums1[q] > nums2[p])
{
k = k - (p - begin2);
size1 = q - begin1 + 1;
size2 -= (p - begin2);
begin2 = p;
}
else
{
if (nums1[q] < nums2[p])
{
k = k - (q - begin1);
size1 -= (q - begin1);
begin1 = q;
size2 = p - begin2 + 1;
}
else
{
return nums1[q];
}
}
return find_kth(nums1, begin1, size1, nums2, begin2, size2, k);
}
迭代
int find_kth(vector<int>& nums1, int begin1, int size1, vector<int>& nums2, int begin2, int size2, int k)
{
while (!(size2 == 1 || size1 == 1 || k == 1))
{
size1 = min(k, size1);//第k大最多只要前k个
size2 = min(k, size2);
double s = k / static_cast<double>(size1 + size2);//对应的比例位置
int q = s*(size1)+begin1;/**/
int p = s*(size2)+begin2;/**/
if (static_cast<int>(s*(size1)) + static_cast<int>(s*(size2)) > k - 1 && (q - begin1) && (p - begin2))
//调节使 k刚好落在p 或 k
{
--p;
--q;
}
if (static_cast<int>(s*(size1)) + static_cast<int>(s*(size2)) < k - 3 && (q - begin1) && (p - begin2))
//调节使 k刚好落在p 或 k
{
++p;
++q;
}
if (nums1[q] > nums2[p])
{
k = k - (p - begin2);
size1 = q - begin1 + 1;
size2 -= (p - begin2);
begin2 = p;
}
else
{
if (nums1[q] < nums2[p])
{
k = k - (q - begin1);
size1 -= (q - begin1);
begin1 = q;
size2 = p - begin2 + 1;
}
else
{
return nums1[q];
}
}
}
if (k == 1)
{
return min(nums1[begin1], nums2[begin2]);
}
if (size1 == 1)
{
if (begin2 + k - 1 < nums2.size())
return min(max(nums1[begin1], nums2[begin2 + k - 2]), nums2[begin2 + k - 1]);
else
return max(nums1[begin1], nums2[begin2 + k - 2]);
}
if (size2 == 1)
{
if (begin1 + k - 1 < nums1.size())
return min(max(nums2[begin2], nums1[begin1 + k - 2]), nums1[begin1 + k - 1]);
else
return max(nums2[begin1], nums1[begin1 + k - 2]);
}
}