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Given two sorted arrays A, B of size m and n respectively. Find the k-th smallest element in the union of A and B. You can assume that there are no duplicate elements.
O(lg m + lg n) solution:
1. Maintaining the invariant
i + j = k - 1,
2. If Bj-1 < Ai < Bj, then Ai must be the k-th smallest,
or else if Ai-1 < Bj < Ai, then Bj must be the k-th smallest.
code:
int findKthSmallest(int A[], int m, int B[], int n, int k) { assert(A && m >= 0 && B && n >= 0 && k > 0 && k <= m+n); int i = (int)((double)m / (m+n) * (k-1)); int j = (k-1) - i; assert(i >= 0 && j >= 0 && i <= m && j <= n); int Ai_1 = ((i == 0) ? INT_MIN : A[i-1]); int Bj_1 = ((j == 0) ? INT_MIN : b[j-1]); int Ai = ((i == m) ? INT_MAX : A[i]); int Bj = ((j == n) ? INT_MAX : B[j]); if (Bj_1 < Ai && Ai < Bj) return Ai; else if (Ai_1 < Bj && Bj < Ai) return Bj; assert((Ai > Bj && Ai_1 > Bi) || (Ai < Bj && Ai < Bj_1)); if (Ai < Bj) return findKthSmallest(A+i+1, m-i-1, B, j, k-i-1); else return findKthSmallest(A, i, B+j+1, n-j-1, k-j-1); }
Find the k-th Smallest Element in the Union of Two Sorted Arrays
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原文地址：http://www.cnblogs.com/litao-tech/p/4217155.html
