#Leetcode# 4. Median of Two Sorted Arrays

https://leetcode.com/problems/median-of-two-sorted-arrays/

 

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)).

You may assume nums1 and nums2 cannot be both empty.

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

代码:

class Solution {
public:
    double findMedianSortedArrays(vector<int>& nums1, vector<int>& nums2) {
        vector<int> v;
        for(int i = 0; i < nums1.size(); i ++)
            v.push_back(nums1[i]);
        for(int i = 0; i < nums2.size(); i ++)
            v.push_back(nums2[i]);
        
        sort(v.begin(), v.end());
        double ans = 0.0;
        if(v.size() % 2 == 0)
            ans = 1.0 * (v[v.size() / 2 - 1] + v[v.size() / 2 + 1 - 1]) / 2;
        else ans = 1.0 * v[(v.size() + 1) / 2 - 1];
        
        return ans;
    }
};

　

二分

class Solution {
public:
    double findMedianSortedArrays(vector<int>& nums1, vector<int>& nums2) {
        int m = nums1.size(), n = nums2.size();
        if (m < n) return findMedianSortedArrays(nums2, nums1);
        if (n == 0) return ((double)nums1[(m - 1) / 2] + (double)nums1[m / 2]) / 2.0;
        int left = 0, right = n * 2;
        while (left <= right) {
            int mid2 = (left + right) / 2;
            int mid1 = m + n - mid2;
            double L1 = mid1 == 0 ? INT_MIN : nums1[(mid1 - 1) / 2];
            double L2 = mid2 == 0 ? INT_MIN : nums2[(mid2 - 1) / 2];
            double R1 = mid1 == m * 2 ? INT_MAX : nums1[mid1 / 2];
            double R2 = mid2 == n * 2 ? INT_MAX : nums2[mid2 / 2];
            if (L1 > R2) left = mid2 + 1;
            else if (L2 > R1) right = mid2 - 1;
            else return (max(L1, L2) + min(R1, R2)) / 2;
        }
        return -1;
    }
};

　　

　

 

第一道 hard 题目 嘻嘻嘻