topic
There are two sorted arrays nums1 and nums2 of size m and n respectively.
Find the midian of the two sorted arrays. The overall run time complexity should be O(log(m+n)).
You may assume nums1 and nums2 cannot be both empty.
Example1：
nums1 = [1, 3]
nums2 = [2]
The median is 2.0
Example2：
nums1 = [1, 2]
nums2 = [3, 4]
The median is (2+3)/2.0 = 2.5
Thinking
Binary search, recursion
Suppose two arrays A and B is greater than the number \ (\ frac {k} { 2} \) number (a number of deficiencies array \ (\ frac {k} { 2} \) a), respectively, Take the first \ (\ frac {k} { 2} \) elements (array labeled $ \ frac {k} {2 } - $ 1) are compared, and there are three cases:
- A[$\frac{k}{2}-$1] \(\;\)< \(\;\) B[$\frac{k}{2}-$1]
- A[$\frac{k}{2}-$1] \(\;\)=\(\;\) B[$\frac{k}{2}-$1]
- A[$\frac{k}{2}-$1] \(\;\)>\(\;\) B[$\frac{k}{2}-$1]
In the first case, if the first A, \ (\ frac {k} { 2} \) elements (array labeled $ \ frac {k} {2 } -1 $) than the section B \ (\ FRAC {k} {2} \) elements (array subscript \ (\ K FRAC {-1} {2} \) ) small, when a and B described later were combined, subscript a from 0 to ($ \ frac {k} {2} - number between \ (1) can not be a large number of k, i.e., a small array of cut, k should be increased if they are equal, is the need to find elements (a k. -1 elements) is larger than the case of the odd-numbered similarly to take less than the middle, an even number of the two numbers dividing by 2.0 to boundary conditions recursive **:. ** 1) wherein one array is empty, returns directly to another array subscript [k-1] number. (See single array Looking k-th element) 2) If k = 1, i.e. to find a minimum value, a direct comparison of the first two elements of the array (array subscript 0) to. 3) If A [\) \ K FRAC} {2} {- \ (. 1] = B [\) \ K FRAC} {2} {- $. 1], only one of them returned.
Tips
Cut the number of k-looking ideas
Cut through, the array is cut into left and right parts: \ (Part Left_ {}, {right_ Part} \) , two elements generated at this time, namely the maximum value of the left \ (L_ {max} \) and the minimum on the right value \ (min R_ {} \) . Cut process may Cut a number, then this number belongs to both the left and the right belongs; Cut may also be in the middle of two numbers.
- Single k-th element array looking
for an ordered array A, in the k number (the array subscript k-1) cut at the return value \ (A [k-. 1] = max {L} \) , namely the number of k, the maximum number of left-hand portion. (k = 1,2, ...) - Double-Array find the first k elements
| \(Left_{part}\) | \(C_i\) | \(Right_{part}\) |
| :---: |: --- :|: --- :|
|\(a_1,a_2,\cdots,a_i\) | / | \(a_{i+1},a_{i+2},\cdots,a_m\)|
|\(b_1,b_2,\cdots,b_j\) | / | \(b_{j+1},b_{j+2},\cdots,a_n\)|
Define (max_1 L_ {} \) \ , \ (max_2 L_ {} \) is after Cut \ (Left_part \) a maximum of 1 second and the array, \ (MIN_1 R_ {} \) , \ (R_ {min_2} \) is after Cut \ (Part right_ {} \) minimum value of 1 and the second array. \ (C_1, C_2 \) is the first and second array of Cut.
If satisfied:
. 1) \ (max_1 L_ {} \; <\; R & lt MIN_1} {\) , \ (max_2 L_ {} \; <\; min_2 R_ {} \) . (This must be satisfied, because an ordered array, must be smaller than the left to the right.)
2) \ (max_1 L_ {} \; <= \; min_2 R_ {} \) , and \ (L_ {max_2} \; <= \; MIN_1 R_ {} \) .
The \ (Left_ {part} \) Full less than \ (Part right_ {} \) . If the number of elements left together exactly equal to k , then the k-th element is \ (Max (L_ {max_1} , L_ {max_2}) \)(See single array to find the k th element). If \ (max_1 L_ {} \;> \; min_2 R_ {} \) , many elements of the array 1 described left (large), the need to reduce \ (C_l \) , the (C_2 \) \ increases. Similarly available (max_2 L_ {} \;> \; MIN_1 R_ {} \) \ , a \ (C_l \) is increased, \ (C_2 \) decreases.
C++
class Solution {
public:
double findMedianSortedArrays(vector<int>& nums1, vector<int>& nums2) {
int n1 = nums1.size();
int n2 = nums2.size();
int total = n1 + n2;
if(total % 2){ //如果数组加起来是奇数
return findKth(nums1, 0, nums2, 0, total / 2 + 1);
}
else{ //如果是偶数
return (findKth(nums1, 0, nums2, 0, total / 2 ) + findKth(nums1, 0, nums2, 0, total / 2 + 1) )/2.0;
}
}
//分割的思想寻找第k个数
double findKth(vector<int>& nums1, int l, vector<int>& nums2, int r, int k){
int n1 = nums1.size();
int n2 = nums2.size();
if(n1 - l > n2 -r )
return findKth(nums2, r, nums1, l, k); //始终保证第一个数组是个数是最少的
if(n1-l == 0)
return nums2[r + k -1];
if(k == 1)
return min(nums1[l], nums2[r]);
int p1 = min(k/2 , n1); //保证在第一个数组内做二分查找。
int p2 = k - p1;
if(nums1[l + p1 -1] < nums2[r + p2 -1]){ //左边
return findKth(nums1, l+p1,nums2, r,k - p1);
}
else if(nums1[l + p1 -1] > nums2[r + p2 -1]){ //左边数组1的个数太大
return findKth(nums1, l,nums2, r+p2,k - p2);
}
else{
return nums1[l+p1-1];
}
}
};python
class Solution(object):
def findKth(self, nums1, nums2,k):
k = int(k)
n1 = len(nums1)
n2 = len(nums2)
if n1 > n2:
return self.findKth(nums2, nums1, k)
if n1 == 0:
return nums2[k - 1]
if k == 1:
return min(nums1[0], nums2[0])
p1 = int(min(k / 2, n1))
p2 = k - p1
if nums1[p1 - 1] <= nums2[p2 - 1]:
return self.findKth(nums1[p1:], nums2, p2)
else:
return self.findKth(nums1, nums2[p2:], p1)
def findMedianSortedArrays(self, nums1, nums2):
"""
:type nums1: List[int]
:type nums2: List[int]
:rtype: float
"""
n1 = len(nums1)
n2 = len(nums2)
total = n1 + n2
if(total % 2):
return self.findKth(nums1, nums2, total /2 + 1)
else:
return (self.findKth(nums1, nums2, total /2 )+ self.findKth(nums1, nums2, total /2 + 1))/2.0 reference
[1] https://blog.csdn.net/hk2291976/article/details/51107778
[2] https://leetcode.com/problems/median-of-two-sorted-arrays/solution/