Leetcode 4. Median of Two Sorted Arrays（Hard）

Source:https://leetcode.com/problems/median-of-two-sorted-arrays/
TIme:2020/11/20*

这题有点东西的，之后有时间要再看看。
Solution里讲得非常清楚了

Description
Given two sorted arrays nums1 and nums2 of size m and n respectively, return the median of the two sorted arrays.

Follow up: The overall run time complexity should be O(log (m+n)).

Example 1:

Input: nums1 = [1,3], nums2 = [2]
Output: 2.00000
Explanation: merged array = [1,2,3] and median is 2.
Example 2:

Input: nums1 = [1,2], nums2 = [3,4]
Output: 2.50000
Explanation: merged array = [1,2,3,4] and median is (2 + 3) / 2 = 2.5.
Example 3:

Input: nums1 = [0,0], nums2 = [0,0]
Output: 0.00000
Example 4:

Input: nums1 = [], nums2 = [1]
Output: 1.00000
Example 5:

Input: nums1 = [2], nums2 = []
Output: 2.00000

Constraints:

扫描二维码关注公众号，回复： 14678973 查看本文章

nums1.length == m
nums2.length == n
0 <= m <= 1000
0 <= n <= 1000
1 <= m + n <= 2000
-106 <= nums1[i], nums2[i] <= 106

Approach 1: Recursive Approach

class Solution {
    
    
public:
    double findMedianSortedArrays(vector<int>& nums1, vector<int>& nums2) {
    
    
        int m = nums1.size();
        int n = nums2.size();
        
        if(m > n)
        {
    
    
            vector<int> temp;
            temp = nums1;
            nums1 = nums2;
            nums2 = temp;
            int tmp = m;
            m = n;
            n = tmp;
        }
        
        int iMin = 0, iMax = m, halfLen = (m + n + 1) / 2;
        while(iMin <= iMax)
        {
    
    
            int i = (iMin + iMax) / 2;
            int j = halfLen - i;
            if(i < iMax && nums2[j - 1] > nums1[i])
                iMin = i + 1;
            else if (i > iMin && nums1[i - 1] > nums2[j])
                iMax = i - 1;
            else
            {
    
    
                int maxLeft = 0;
                if(i == 0)
                    maxLeft = nums2[j - 1];
                else if(j == 0)
                    maxLeft = nums1[i - 1];
                else
                    maxLeft = max(nums1[i - 1], nums2[j - 1]);
                if((m + n) % 2 == 1)
                    return maxLeft;
                
                int minRight = 0;
                if(i == m)
                    minRight = nums2[j];
                else if(j == n)
                    minRight = nums1[i];
                else
                    minRight = min(nums1[i], nums2[j]);
                
                return (maxLeft + minRight) / 2.0;
            }
        }
        return 0.0;
    }
};