转自leetcode-cpp-13类from清华大佬的PDF文档
倒是想到上面思路了,但是递归分段的话没想到能直接用迭代器来直接做到简便(之前都是加左右边界条件来做,而且K也需要计算)。
// LeetCode, Median of Two Sorted Arrays
class Solution {
public:
double findMedianSortedArrays(const vector<int>& A, const vector<int>& B) {
const int m = A.size();
const int n = B.size();
int total = m + n;
if (total & 0x1)
return find_kth(A.begin(), m, B.begin(), n, total / 2 + 1);
else
return (find_kth(A.begin(), m, B.begin(), n, total / 2)
+ find_kth(A.begin(), m, B.begin(), n, total / 2 + 1)) / 2.0;
}
private:
static int find_kth(std::vector<int>::const_iterator A, int m,
std::vector<int>::const_iterator B, int n, int k) {
//always assume that m is equal or smaller than n
if (m > n) return find_kth(B, n, A, m, k);
if (m == 0) return *(B + k - 1);
if (k == 1) return min(*A, *B);
//divide k into two parts
int ia = min(k / 2, m), ib = k - ia;
if (*(A + ia - 1) < *(B + ib - 1))
return find_kth(A + ia, m - ia, B, n, k - ia);
else if (*(A + ia - 1) > *(B + ib - 1))
return find_kth(A, m, B + ib, n - ib, k - ib);
else
return A[ia - 1];
}
};