4. Median of Two Sorted Arrays（divide and conquer）

1. Description

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)).

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

2. Analysis

归并排序的时间复杂度为O(log(m+n))。

3. code

class Solution {
public:
static vector<int> merge_sort(vector<int>& nums1, vector<int>& nums2) {
int i = 0, j = 0;
vector<int> res;
while(i < nums1.size() && j < nums2.size()) {
if(nums1[i] <= nums2[j]) {
res.push_back(nums1[i]);
i++;
} else {
res.push_back(nums2[j]);
j++;
}
}

while(i < nums1.size()) {
res.push_back(nums1[i]);
i++;
}
while(j < nums2.size()) {
res.push_back(nums2[j]);
j++;
}
return res;
}

double findMedianSortedArrays(vector<int>& nums1, vector<int>& nums2) {
vector<int> res = merge_sort(nums1, nums2);
int size = res.size();
if(size % 2 == 0) {
return (double)(res[size/2] + res[size/2-1])/2;
} else {
return (double)(res[size/2]);
}
}
};

4. improvement

<待续>