【经典】Median of Two Sorted Arrays
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题目：leetcode
Median of Two Sorted Arrays
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)).
注意：当m+n为偶数时，此题的中位数是中间两个数的平均值。
分析：
假设数组A长度为m，数组B长度为n。
1、令数组A始终为长度小的那个数组。
2、考虑A为空，B不为空的情况。
3、m+n为偶数和奇数的情况要分开讨论。
4、此题有两种思路：一种思路是求两个有序数组的第k个数，这里k=(m+n)/2. 这种方法的时间复杂度是O(log(m+n)). 另一种思路是首先单独处理m==1 和 m==2的情况。m>=3时，每次递归都在中位数的前后删除deletenum长度的数字（这个deletenum约等于数组A剩余长度的一半），直到数组A长度为2（这是递归的终止条件）。第二种思路的时间复杂度为O(log(min(m,n))).
思路1：时间复杂度O(log(m+n))
class Solution {public: double findMedianSortedArrays(int A[], int m, int B[], int n) { if(m>n) return findMedianSortedArrays(B,n,A,m); if(m==0) { if(n==0) throw exception(); if(n & 1) { return B[(n-1)/2]; } else { double m1=B[(n-1)/2],m2=B[(n-1)/2+1]; return (m1+m2)/2; } } int total=m+n; if(total & 1) return find_kth(A,0,m-1,B,0,n-1,total/2); else return (find_kth(A,0,m-1,B,0,n-1,total/2)+find_kth(A,0,m-1,B,0,n-1,total/2-1))/2.0; } double find_kth(int A[], int abegin,int aend, int B[], int bbegin,int bend,int index) { if(abegin>aend) { return B[bbegin+index]; } if(bbegin>bend) { return A[abegin+index]; } int len1=aend-abegin+1,len2=bend-bbegin+1; if(len1>len2) return find_kth(B,bbegin,bend,A,abegin,aend,index); int ia=min(index/2+abegin,aend),ib=index-(ia-abegin+1)+bbegin; if(index==0) return min(A[abegin],B[bbegin]); else if(A[ia]==B[ib]) return A[ia]; else if(A[ia]<B[ib]) { return find_kth(A,ia+1,aend,B,bbegin,bend,index-(ia-abegin+1)); } else if(A[ia]>B[ib]) { return find_kth(A,abegin,aend,B,ib+1,bend,index-(ib-bbegin+1)); } } };思路2：时间复杂度O(log(min(m,n)))
class Solution {public: double findMedianSortedArrays(int A[], int m, int B[], int n) { if(m>n) return findMedianSortedArrays(B,n,A,m); if(m==0) { if(n==0) throw exception(); if(n & 1) { return B[(n-1)/2]; } else { double m1=B[(n-1)/2],m2=B[(n-1)/2+1]; return (m1+m2)/2; } } int total=m+n; if(total & 1) return findMedianSortedArrays_up(A,0,m-1,B,0,n-1); else return ( findMedianSortedArrays_down(A,0,m-1,B,0,n-1)+findMedianSortedArrays_up(A,0,m-1,B,0,n-1) )/2.0; } //比较小的那个数作为中位数 double findMedianSortedArrays_up(int A[], int al,int ar, int B[], int bl,int br) { int lena=ar-al+1,lenb=br-bl+1; int mida=al+(ar-al)/2,midb=bl+(br-bl)/2; if(lena==1 || lena==2 ) { if(lena==1) return one_multiple_up(B,bl,br,A[al]); else { return two_multiple_up(A,al,ar,B,bl,br); } } else { int deletenum=lena&1?lena/2:lena/2-1; if( A[mida]==B[midb]) { return A[mida]; } else if( A[mida]<B[midb]) { return findMedianSortedArrays_up(A,mida,ar,B,bl,br-deletenum); } else { int offset=lena&1?0:1; return findMedianSortedArrays_up(A,al,mida+offset,B,bl+deletenum,br); } } } double one_multiple_up(int num[], int l,int r,int number) { int mid=l+(r-l)/2,len=r-l+1; if(num[mid]==number) { return num[mid]; } else if(num[mid]<number) { if(len & 1) { return num[mid]; } else { return min(number,num[mid+1]); } } else { if(len & 1) { if(len==1) { return number; } else { return max(number,num[mid-1]); } } else { return num[mid]; } } } double two_multiple_up(int A[], int al,int ar, int B[], int bl,int br) { int mida=al+(ar-al)/2,midb=bl+(br-bl)/2; int lenb=br-bl+1; if((A[al]==A[ar] && A[al]==B[midb]) || (A[al]<B[midb] && A[ar]>B[midb])) { return B[midb]; } else if(A[al]==B[midb] || A[ar]==B[midb]) { return B[midb]; } else if(A[ar]<B[midb]) { if(lenb &1) { return max(A[ar],B[midb-1]);/////// } else { if(lenb==2) return A[ar]; else { return max(B[midb-1],A[ar]); } } } else//A[al]>B[midb] { return min(B[midb+1],A[al]); } } ////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////////// //比较大的那个数作为中位数 double findMedianSortedArrays_down(int A[], int al,int ar, int B[], int bl,int br) { int lena=ar-al+1,lenb=br-bl+1; int mida=al+(ar-al)/2,midb=bl+(br-bl)/2; if((lena & 1) == 0) { mida++; } if((lenb & 1) == 0) { midb++; } if(lena==1 || lena==2) { if(lena==1) return one_multiple_down(B,bl,br,A[al]); else { return two_multiple_down(A,al,ar,B,bl,br); } } else { int deletenum=lena&1?lena/2:lena/2-1; if( A[mida]==B[midb]) { return A[mida]; } else if( A[mida]<B[midb]) { int offset=lena&1?0:1; return findMedianSortedArrays_down(A,mida-offset,ar,B,bl,br-deletenum); } else { //int offset=lena&1?0:1; return findMedianSortedArrays_down(A,al,mida,B,bl+deletenum,br); } } } double one_multiple_down(int num[], int l,int r,int number) { int mid=l+(r-l)/2,len=r-l+1; if((len & 1) == 0) { mid++; } if(num[mid]==number) { return num[mid]; } else if(num[mid]<number) { if(len & 1) { if(len==1) { return number; } else { return min(number,num[mid+1]); } } else { return num[mid]; } } else { if(len & 1) { return num[mid]; } else { return min(number,num[mid-1]); } } } double two_multiple_down(int A[], int al,int ar, int B[], int bl,int br) { int lena=ar-al+1,lenb=br-bl+1; int mida=al+(ar-al)/2,midb=bl+(br-bl)/2; if((lena & 1) == 0) { mida++; } if((lenb & 1) == 0) { midb++; } if((A[al]==A[ar] && A[al]==B[midb]) || (A[al]<B[midb] && A[ar]>B[midb])) { return B[midb]; } else if(A[ar]<=B[midb]) { return max(A[ar],B[midb-1]); } else if(A[al]>=B[midb]) { if(lenb &1) { return min(B[midb+1],A[al]); } else { if(lenb==2) { return A[al]; } else { return min(B[midb+1],A[al]); } } } }}; 0 0
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