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

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2. Analysis

归并排序的时间复杂度为O(log(m+n))。

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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>