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class Solution {//more detail refer to: http://fisherlei.blogspot.com/2012/12/leetcode-median-of-two-sorted-arrays.html//using the method of getting the kth number in the two sorted array to solve the median problem//divide-and-conquer//very clean and concisepublic: double findMedianSortedArrays(int A[], int m, int B[], int n) { if((n+m)%2 ==0) { return (GetMedian(A,m,B,n, (m+n)/2) + GetMedian(A,m,B,n, (m+n)/2+1))/2.0; } else return GetMedian(A,m,B,n, (m+n)/2+1); } int GetMedian(int a[], int n, int b[], int m, int k)//get the kth number in the two sorted array { //assert(a && b); if (n <= 0) return b[k-1]; if (m <= 0) return a[k-1]; if (k <= 1) return min(a[0], b[0]); //a: section1 section2 //b: section3 section4 if (b[m/2] >= a[n/2]) { if ((n/2 + 1 + m/2) >= k) return GetMedian(a, n, b, m/2, k);//abort section 4 else return GetMedian(a + n/2 + 1, n - (n/2 + 1), b, m, k - (n/2 + 1)); //abort section 1 } else { if ((m/2 + 1 + n/2) >= k) return GetMedian( a, n/2,b, m, k);//abort section 2 else return GetMedian( a, n, b + m/2 + 1, m - (m/2 + 1),k - (m/2 + 1));//abort section 3 } } };
second time
class Solution {public: int GetKthNumber(int A[], int m, int B[], int n, int k) { //terminate case if(m <= 0) return B[k]; if(n <= 0) return A[k]; if(k <= 0) return min(A[0], B[0]); //recursion //A:s1,median,s2 //B:s3,median,s4 if(A[m/2] > B[n/2])//s2>s3 { if(m/2+n/2+1 >= k+1) { //A:s1 //B:s3,median,s4 GetKthNumber(A, m/2, B, n, k); } else { //A:s1,median,s2 //B:s4 GetKthNumber(A, m, B+(n/2+1), n-(n/2+1), k-(n/2+1)); } } else//A[m/2] < B[n/2] { if(m/2+n/2+1 >= k+1) GetKthNumber(A, m, B, n/2, k); else GetKthNumber(A+(m/2+1), m-(m/2+1), B, n, k-(m/2+1)); } } double findMedianSortedArrays(int A[], int m, int B[], int n) { // Start typing your C/C++ solution below // DO NOT write int main() function if((m+n)%2 == 0)//even return (double)(GetKthNumber(A, m, B, n, (m+n)/2-1)+GetKthNumber(A, m, B, n, (m+n)/2))/2.0; else return (double)GetKthNumber(A, m, B, n, (m+n)/2); }};
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