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002* Median of the Sorted Arrays

There is sorted arrays A and B of size m and N respectively. Find The median of the sorted arrays. The overall run time complexity should be O (log (m+n)).

1 classSolution:2     deffindkth (self, A, A_start, B, B_start, K):3         ifA_start >=Len (A):4             returnB[b_start + k-1]5         ifB_start >=Len (B):6             returnA[a_start + k-1]7         ifK = = 1:8             returnmin (A[a_start], B[b_start])9A_key = 0x3f3f3f3fTen         ifA_start + K/2-1 <Len (A): OneA_key = A[a_start + K/2-1] AB_key = 0x3f3f3f3f -         ifB_start + K/2-1 <Len (B): -B_key = B[b_start + K/2-1] the         ifA_key <B_key: -             returnSelf.findkth (A, A_start + K/2, B, B_start, K-k/2) -         Else: -             returnSelf.findkth (A, A_start, B, B_start + K/2, K-k/2) +     #@return A float -     deffindmediansortedarrays (self, A, B): +n = Len (A) +Len (B) A         ifn% 2 = =0: at             return(Self.findkth (A, 0, b, 0, N/2) + self.findkth (A, 0, b, 0, N/2 + 1)/2.0 -         Else: -             returnSelf.findkth (A, 0, B, 0, N/2 + 1)

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Introduction to the algorithm 9.3-8, set X[1..N] and Y[1..N] is two arrays, each containing n ordered number. An O (LGN) time algorithm is given for calculating the median of all 2n elements in the array x and Y.

1 intFindmedian (intA[],intB[],intNintLowintHigh ) {  2     if(Low > High)returnNot_found; 3     Else {  4         intK = (Low+high)/2; 5         if(K==n && a[n]<=b[1])returnA[n]; 6         Else if(K<n && b[n-k]<=a[k] && a[k]<=b[n-k+1])returnA[k]; 7         Else if(A[k] > b[n-k+1])returnFindmedian (a,b,n,low,k-1); 8         Else returnFindmedian (a,b,n,k+1, high); 9     }  Ten }   One intTwoarraymedian (intX[],intY[],intN) { A     intMedian = Findmedian (X,y,n,1, N);  -     if(median==not_found) median = Findmedian (Y,x,n,1, N);  -     returnmedian;  the}

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We're going to take the median, and notice that a number x is the median, and if and only if the number of n-1 is smaller than x, and there is a number of n larger than X, we first assume that the median is in array a, where the median number of squares may be.
The median interval is assumed to be [l,r],k = (l+r)/2. If A[K] is the median, the number of k-1 in array A is smaller than a[k], and the n-k number is greater than a[k]. If the number of n-k in B is smaller than a[k], there is a number of k greater than a[k], then a[k] is the median.
Because B is ordered, so b[n-k]<=a[k]<=b[n-k+1].
If a[k]>b[n-k+1], then a number smaller than a[k] is at least N, so the median is less than a[k], so it is in the interval [l,k-1].
The inverse is in the interval [k+1,r].
If the l>r, then the median is not in a, the B array for the same two-point operation can be.

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