LeetCode-4. Median of Sorted Arrays

Topic:

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)).

Test instructions

There are two sets of sorted arrays nums1 and NUMS2, with lengths of M and N, to find the median of two sets of numbers, with the time complexity of O (log (m+n)).

The median value is: for an odd-length array such as {1,3,4}, the median value is 3, and for an even-numbered array {1,2,5,7}, the median is (2+5)/2.0=3.5.

Problem-solving ideas (borrowed from the discussion hot first answer ideas, really admire this person will algorithm problem with mathematical thinking analysis):

1. For sorted array A of length m, we can use subscript I to divide it into two groups: {a[0], ... A[i-1], | A[i], ... A[m]}.

A total of m+1 species (i = 0~m). And the left I element of the array is no more than a[i], the m-i element on the right is not less than a[i].

Similarly, a sorted array B with length n is split by J, and A and B are merged by the i + J to create an array of result.

2. To get the median value, you need to find the corresponding I and J to meet the following two conditions:

(1) i + j = m-i + n-j (M+n is an even number case)

or i + j = m-i + n-j + 1 (m+n is an odd case);

(2) All values of a[0] ~ A[i-1] and b[0] ~ B[j-1] are not all values of a[i] ~ A[m] and b[j] ~ B [n].

3. The above two conditions are equivalent to:

(1) Assuming n > m,j = (m + n + 1)/2-i;

(2) A[i-1] < B[j] and b[j-1] < a[i].

Also consider the case where the boundary value i = 0 or m, j = 0 or n is required.

4. [Coding ideas] Next, you can use the binary lookup method to help obtain the median value (the time complexity of the topic can be associated with the binary search method).

(1) Set imin = 0, Imax = m, start searching in (Imin, IMAX);

(2) Set i = (Imin + IMAX)/2, half = (m + n + 1)/2, j = half-i;

(3) if (A[i-1] > b[j]) imax = i-1; Search area narrowed to I left half

else if (A[i] < b[j-1]) Imin = i + 1; Search area narrowed to I right half

else break;

(4) if (0==i) NUM1 = b[j-1]; I==0 j = half, i.e. b[j-1] is the maximum left of the merged Result

else if (0==j) NUM1 = a[i-1]; J==0 A[i-1] is the maximum value left for Result

else NUM1 = max (b[j-1], a[i-1]); In other cases, the left maximum value of Result is taken

M+n is an odd number, return to NUM1 directly

if ((m+n)%2==1)

return NUM1;

if (m==i) num2 = B[j]; I==m, when j=0 NUM1 = A[i-1] is introduced, it is necessary to take num2 = b[j at this time)

else if (n = = j) num2 = A[i]; J==n num2 = A[i]

else num2 = min (B[j], a[i]); In other cases, take the minimum value of the right part of the Result

Return ((NUM1 + num2)/2.0);

C # code:

 Public classsolution{ Public DoubleFindmediansortedarrays (int[] nums1,int[] nums2) {        intm = nums1. Length, n =Nums2.        Length; if(M > N)returnfindmediansortedarrays (NUMS2, NUMS1); intImin =0, Imax = m, half = (m + n +1) /2; inti =0, j =0;  while(Imin <=IMAX) {i= (Imin + IMAX)/2; J= Half-i; if(I >0&& J < n && Nums1[i-1] > Nums2[j]) imax = i-1; Else if(J >0&& i < m && Nums2[j-1] > Nums1[i]) imin = i +1; Else  Break; }        DoubleNUM1 =0, num2 =0; if(0= = i) Num1 = Nums2[j-1]; Else if(0= = j) Num1 = Nums1[i-1]; ElseNUM1 = Math.max (Nums2[j-1], Nums1[i-1]); if((m + N)%2==1)            returnNUM1; if(m = = i) num2 =Nums2[j]; Else if(n = = j) num2 =Nums1[i]; Elsenum2 =math.min (Nums2[j], nums1[i]); return((NUM1 + num2)/2); }}

　　　　　　

　　

　　

LeetCode-4. Median of Sorted Arrays