计算两个数组的中位数
Question
There are two sorted arrays nums1 and nums2 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)).
前提条件
两个数组已排序
Contents
- Same Size Arrays - Median of Two-Same-Size-Sorted-Arrays
- Different Size Arrays - Median of Two-Different-Size-Arrays
- Both - combine both same size and different size
Same Size Arrays
Example:
nums1 = [1, 2]
nums2 = [3, 4]
The median is (2 + 3)/2 = 2.5
中位数
Median: In probability theory and statistics, a median is described as the number separating the higher half of a sample, a population, or a probability distribution, from the lower half.
The median of a finite list of numbers can be found by arranging all the numbers from lowest value to highest value and picking the middle one.
Method
1. Method 1 (Simply count while Merging)
Time Complexity: O(n)
Use merge procedure of merge sort. Keep track of count while comparing elements of two arrays. If count becomes n(For 2n elements), we have reached the median. Take the average of the elements at indexes n-1 and n in the merged array. See the below implementation.
2. Method 2 (By comparing the medians of two arrays)
Time Complexity: O(logn)
This method works by first getting medians of the two sorted arrays and then comparing them.
Let ar1 and ar2 be the input arrays.
Algorithm:
1) Calculate the medians m1 and m2 of the input arrays ar1[]
and ar2[] respectively.
2) If m1 and m2 both are equal then we are done.
return m1 (or m2)
3) If m1 is greater than m2, then median is present in one
of the below two subarrays.
a) From first element of ar1 to m1 (ar1[0...|_n/2_|])
b) From m2 to last element of ar2 (ar2[|_n/2_|...n-1])
4) If m2 is greater than m1, then median is present in one
of the below two subarrays.
a) From m1 to last element of ar1 (ar1[|_n/2_|...n-1])
b) From first element of ar2 to m2 (ar2[0...|_n/2_|])
5) Repeat the above process until size of both the subarrays
becomes 2.
6) If size of the two arrays is 2 then use below formula to get
the median.
Median = (max(ar1[0], ar2[0]) + min(ar1[1], ar2[1]))/2
Example:
ar1[] = {1, 12, 15, 26, 38}
ar2[] = {2, 13, 17, 30, 45}
For above two arrays m1 = 15 and m2 = 17
For the above ar1[] and ar2[], m1 is smaller than m2. So median is present in one of the following two subarrays.
[15, 26, 38] and [2, 13, 17]
Let us repeat the process for above two subarrays:
m1 = 26 m2 = 13.
m1 is greater than m2. So the subarrays become
[15, 26] and [13, 17]
Now size is 2, so median = (max(ar1[0], ar2[0]) + min(ar1[1], ar2[1]))/2
= (max(15, 13) + min(26, 17))/2
= (15 + 17)/2
= 16
Different Size Arrays
Time Complexity: O(LogM + LogN)
Example:
nums1 = [1, 3]
nums2 = [2]
The median is 2.0
The approach discussed in this post is similar to method 2 of equal size post. The basic idea is same, we find the median of two arrays and compare the medians to discard almost half of the elements in both arrays. Since the number of elements may differ here, there are many base cases that need to be handled separately. Before we proceed to complete solution, let us first talk about all base cases.
Let the two arrays be A[N] and B[M]. In the following explanation, it is assumed that N is smaller than or equal to M.
Base cases:
The smaller array has only one element
Case 0: N = 0, M = 2
Case 1: N = 1, M = 1.
Case 2: N = 1, M is odd
Case 3: N = 1, M is even
The smaller array has only two elements
Case 4: N = 2, M = 2
Case 5: N = 2, M is odd
Case 6: N = 2, M is even
Case 0: There are no elements in first array, return median of second array. If second array is also empty, return -1.
Case 1: There is only one element in both arrays, so output the average of A[0] and B[0].
Case 2: N = 1, M is odd
Let B[5] = {5, 10, 12, 15, 20}
First find the middle element of B[], which is 12 for above array. There are following 4 sub-cases.
…2.1 If A[0] is smaller than 10, the median is average of 10 and 12.
…2.2 If A[0] lies between 10 and 12, the median is average of A[0] and 12.
…2.3 If A[0] lies between 12 and 15, the median is average of 12 and A[0].
…2.4 If A[0] is greater than 15, the median is average of 12 and 15. In all the sub-cases, we find that 12 is fixed. So, we need to find the median of B[ M / 2 – 1 ], B[ M / 2 + 1], A[ 0 ] and take its average with B[ M / 2 ].
Case 3: N = 1, M is even
Let B[4] = {5, 10, 12, 15}
First find the middle items in B[], which are 10 and 12 in above example. There are following 3 sub-cases.
…3.1 If A[0] is smaller than 10, the median is 10.
…3.2 If A[0] lies between 10 and 12, the median is A[0].
…3.3 If A[0] is greater than 12, the median is 12. So, in this case, find the median of three elements B[ M / 2 – 1 ], B[ M / 2] and A[ 0 ].
Case 4: N = 2, M = 2 There are four elements in total. So we find the median of 4 elements.
Case 5: N = 2, M is odd Let B[5] = {5, 10, 12, 15, 20} The median is given by median of following three elements: B[M/2], max(A[0], B[M/2 – 1]), min(A[1], B[M/2 + 1]).
Case 6: N = 2, M is even
Let B[4] = {5, 10, 12, 15}
The median is given by median of following four elements: B[M/2], B[M/2 – 1], max(A[0], B[M/2 – 2]), min(A[1], B[M/2 + 1])
Remaining Cases:
Once we have handled the above base cases, following is the remaining process.
1) Find the middle item of A[] and middle item of B[].
…..1.1) If the middle item of A[] is greater than middle item of B[], ignore the last half of A[], let length of ignored part is idx. Also, cut down B[] by idx from the start.
…..1.2) else, ignore the first half of A[], let length of ignored part is idx. Also, cut down B[] by idx from the last.
Both
传送门：
参考文献
- GeeksforGeeks - median-of-two-sorted-arrays
- GeeksforGeeks - median-of-two-sorted-arrays-of-different-sizes