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)).
解题思路:
该题可以解决所有求有序数组A和B有序合并之后第k小的数!
该题的重要结论:
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。
具体的分析过程可以参考http://blog.csdn.net/zxzxy1988/article/details/8587244
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; } } };