4. Median of Sorted Arrays

There is sorted arrays nums1 and nums2 of size M and N respectively.

Find The median of the sorted arrays. The overall run time complexity should be O (log (m+n)).

Example 1:

NUMS1 = [1, 3]nums2 = [2]the median is 2.0

Example 2:

NUMS1 = [1, 2]NUMS2 = [3, 4]the median is (2 + 3)/2 = 2.5

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Find the median of two arrays,

The definition of the median is clear: the ordered array, array.size=n is an odd number, (n+1)/2 digits; otherwise it is the average of N/2 and (n+1)/2.

========== ideas:

Division of Ideas,

The first step: follow the merge idea to find two in the array of K-large

The second step is to return the median by the method of calculating the median number.

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Code implementation,

classSolution { Public:    DoubleFindmediansortedarrays (vector<int>& Nums1, vector<int>&nums2) {        intLengtha =nums1.size (); intLENGTHB =nums2.size (); intTotal = lengtha+LENGTHB; if(Total &0x1){            returnFind_kth (nums1,nums2,total/2+1); }Else{            return(Find_kth (nums1,nums2,total/2)+find_kth (Nums1,nums2,total/2+1))/2.0; }    }Private:    intFind_kth (vector<int> &A,vector<int> &b,intk) {std::vector<int>::const_iterator P1 =A.begin (); Std::vector<int>::const_iterator P2 =B.begin (); intm =0;  while(P1!=a.end () && p2!=B.end ()) {            if(*p1<=*p2 && m== (K-1)){                return*P1; }Else if(*p1>*p2 && m== (K-1)){                return*P2; }            if(*p1<=*p2) P1++; ElseP2++; M++; }// while         while(p1!=A.end ()) {            if(m== (K-1)){                return*P1; } p1++; M++; }         while(p2!=B.end ()) {            if(m== (K-1)){                return*P2; } P2++; M++; }        return-1; }};

4. Median of Sorted Arrays